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Commitef7e407

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author
liningrui
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add new solutions
1 parent579ed13 commitef7e407

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‎script/[100]相同的树.py

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# 给你两棵二叉树的根节点 p 和 q ，编写一个函数来检验这两棵树是否相同。
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#
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# 如果两个树在结构上相同，并且节点具有相同的值，则认为它们是相同的。
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#
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#
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#
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# 示例 1：
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#
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#
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# 输入：p = [1,2,3], q = [1,2,3]
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# 输出：true
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#
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#
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# 示例 2：
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#
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#
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# 输入：p = [1,2], q = [1,null,2]
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# 输出：false
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#
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#
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# 示例 3：
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#
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#
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# 输入：p = [1,2,1], q = [1,1,2]
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# 输出：false
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#
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#
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#
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#
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# 提示：
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#
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#
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# 两棵树上的节点数目都在范围 [0, 100] 内
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# -10⁴ <= Node.val <= 10⁴
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#
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# Related Topics 树 深度优先搜索 广度优先搜索 二叉树 👍 862 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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classSolution:
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defisSameTree(self,p:TreeNode,q:TreeNode)->bool:
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defhelper(p,q):
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ifnotpandnotq:
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returnTrue
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elifnotp:
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returnFalse
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elifnotq:
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returnFalse
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elifp.val!=q.val:
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returnFalse
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returnhelper(p.right,q.right)andhelper(p.left,q.left)
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returnhelper(p,q)
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# leetcode submit region end(Prohibit modification and deletion)

‎script/[101]对称二叉树.py

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# 给你一个二叉树的根节点 root ， 检查它是否轴对称。
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#
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#
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#
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# 示例 1：
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#
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#
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# 输入：root = [1,2,2,3,4,4,3]
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# 输出：true
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#
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#
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# 示例 2：
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#
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#
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# 输入：root = [1,2,2,null,3,null,3]
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# 输出：false
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#
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#
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#
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#
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# 提示：
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#
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#
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# 树中节点数目在范围 [1, 1000] 内
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# -100 <= Node.val <= 100
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#
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#
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#
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#
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# 进阶：你可以运用递归和迭代两种方法解决这个问题吗？
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# Related Topics 树 深度优先搜索 广度优先搜索 二叉树 👍 2013 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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classSolution:
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defisSymmetric(self,root:Optional[TreeNode])->bool:
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defhelper(p,q):
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ifnotpandnotq:
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returnTrue
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elifnotp:
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returnFalse
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elifnotq:
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returnFalse
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elifp.val!=q.val:
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returnFalse
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returnhelper(p.left,q.right)andhelper(p.right,q.left)
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returnhelper(root.left,root.right)
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# leetcode submit region end(Prohibit modification and deletion)
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# 给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。
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#
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#
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#
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# 示例 1：
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#
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#
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# 输入：root = [3,9,20,null,null,15,7]
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# 输出：[[3],[9,20],[15,7]]
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#
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#
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# 示例 2：
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#
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#
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# 输入：root = [1]
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# 输出：[[1]]
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#
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#
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# 示例 3：
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#
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#
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# 输入：root = []
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# 输出：[]
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#
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#
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#
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#
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# 提示：
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#
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#
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# 树中节点数目在范围 [0, 2000] 内
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# -1000 <= Node.val <= 1000
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#
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# Related Topics 树 广度优先搜索 二叉树 👍 1391 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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classSolution:
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deflevelOrder(self,root:TreeNode)->List[List[int]]:
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ifnotroot:
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return []
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result= []
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stack= [root]
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whilestack:
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l=len(stack)
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tmp= []
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foriinrange(l):
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node=stack.pop(0)#注意是pop left
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tmp.append(node.val)
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ifnode.left:
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stack.append(node.left)
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ifnode.right:
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stack.append(node.right)
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result.append(tmp)
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returnresult
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# leetcode submit region end(Prohibit modification and deletion)
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# 给你二叉树的根节点 root ，返回其节点值的 锯齿形层序遍历 。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。
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#
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#
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#
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# 示例 1：
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#
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#
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# 输入：root = [3,9,20,null,null,15,7]
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# 输出：[[3],[20,9],[15,7]]
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#
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#
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# 示例 2：
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#
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#
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# 输入：root = [1]
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# 输出：[[1]]
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#
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#
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# 示例 3：
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#
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#
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# 输入：root = []
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# 输出：[]
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#
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#
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#
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#
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# 提示：
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#
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#
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# 树中节点数目在范围 [0, 2000] 内
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# -100 <= Node.val <= 100
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#
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# Related Topics 树 广度优先搜索 二叉树 👍 668 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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classSolution:
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defzigzagLevelOrder(self,root:TreeNode)->List[List[int]]:
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ifnotroot:
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return []
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result= []
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stack= [root]
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flag=1
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whilestack:
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l=len(stack)
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tmp= []
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foriinrange(l):
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node=stack.pop(0)#注意是pop left
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tmp.append(node.val)
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ifnode.left:
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stack.append(node.left)
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ifnode.right:
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stack.append(node.right)
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ifflag>0:
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result.append(tmp)
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else:
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result.append(tmp[::-1])
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flag*=-1
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returnresult
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# leetcode submit region end(Prohibit modification and deletion)
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# 给定一个二叉树，找出其最大深度。
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#
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# 二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。
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#
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# 说明: 叶子节点是指没有子节点的节点。
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#
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# 示例：
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# 给定二叉树 [3,9,20,null,null,15,7]，
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#
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# 3
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# / \
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# 9 20
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# / \
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# 15 7
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#
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# 返回它的最大深度 3 。
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# Related Topics 树 深度优先搜索 广度优先搜索 二叉树 👍 1300 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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classSolution:
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defmaxDepth(self,root:Optional[TreeNode])->int:
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defdfs(root):
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ifnotroot:
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return0
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left=dfs(root.left)
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right=dfs(root.right)
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returnmax(left,right)+1
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returndfs(root)
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# leetcode submit region end(Prohibit modification and deletion)
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# 给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并
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# 返回其根节点。
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#
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#
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#
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# 示例 1:
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#
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#
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# 输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
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# 输出: [3,9,20,null,null,15,7]
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#
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#
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# 示例 2:
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#
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#
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# 输入: preorder = [-1], inorder = [-1]
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# 输出: [-1]
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#
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#
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#
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#
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# 提示:
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#
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#
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# 1 <= preorder.length <= 3000
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# inorder.length == preorder.length
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# -3000 <= preorder[i], inorder[i] <= 3000
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# preorder 和 inorder 均 无重复 元素
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# inorder 均出现在 preorder
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# preorder 保证 为二叉树的前序遍历序列
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# inorder 保证 为二叉树的中序遍历序列
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#
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# Related Topics 树 数组 哈希表 分治 二叉树 👍 1663 👎 0
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# leetcode submit region begin(Prohibit modification and deletion)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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classSolution:
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defbuildTree(self,preorder:List[int],inorder:List[int])->TreeNode:
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defdfs(left,right):
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ifleft>right:
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returnNone
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val=preorder.pop(0)
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root=TreeNode(val)
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pos=inorder.index(val)
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root.left=dfs(left,pos-1)
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root.right=dfs(pos+1,right)
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returnroot
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returndfs(0,len(preorder)-1)
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# leetcode submit region end(Prohibit modification and deletion)

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