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Commita060f63

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[LEET-545] add 545
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‎leetcode-algorithms/README.md

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##Algorithms
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| # | Title | Solutions | Time | Space | Difficulty | Tag | Notes
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|-----|----------------|---------------|---------------|---------------|-------------|--------------|-----
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|545|[Boundary of Binary Tree](https://leetcode.com/problems/boundary-of-binary-tree/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/BoundaryofBinaryTree.java) | O(n) |O(n) | Medium | Recursion
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|544|[Output Contest Matches](https://leetcode.com/problems/output-contest-matches/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/OutputContestMatches.java) | O(n) |O(n) | Medium | Recursion
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|543|[Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/DiameterofBinaryTree.java) | O(n) |O(h) | Easy | Tree/DFS/Recursion
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|542|[01 Matrix](https://leetcode.com/problems/01-matrix/)|[Solution](../../master/leetcode-algorithms/src/main/java/com/stevesun/solutions/_01Matrix.java) | O(m*n) |O(n) | Medium | BFS
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packagecom.stevesun.solutions;
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importcom.stevesun.common.classes.TreeNode;
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importjava.util.ArrayList;
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importjava.util.HashSet;
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importjava.util.List;
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importjava.util.Set;
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/**
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* Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes.
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Left boundary is defined as the path from root to the left-most node. Right boundary is defined as the path from root to the right-most node. If the root doesn't have left subtree or right subtree, then the root itself is left boundary or right boundary. Note this definition only applies to the input binary tree, and not applies to any subtrees.
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The left-most node is defined as a leaf node you could reach when you always firstly travel to the left subtree if exists. If not, travel to the right subtree. Repeat until you reach a leaf node.
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The right-most node is also defined by the same way with left and right exchanged.
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Example 1
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Input:
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1
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\
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2
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/ \
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3 4
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Ouput:
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[1, 3, 4, 2]
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Explanation:
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The root doesn't have left subtree, so the root itself is left boundary.
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The leaves are node 3 and 4.
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The right boundary are node 1,2,4. Note the anti-clockwise direction means you should output reversed right boundary.
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So order them in anti-clockwise without duplicates and we have [1,3,4,2].
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Example 2
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Input:
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____1_____
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/ \
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2 3
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/ \ /
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4 5 6
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/ \ / \
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7 8 9 10
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Ouput:
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[1,2,4,7,8,9,10,6,3]
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Explanation:
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The left boundary are node 1,2,4. (4 is the left-most node according to definition)
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The leaves are node 4,7,8,9,10.
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The right boundary are node 1,3,6,10. (10 is the right-most node).
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So order them in anti-clockwise without duplicate nodes we have [1,2,4,7,8,9,10,6,3].
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*/
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publicclassBoundaryofBinaryTree {
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publicList<Integer>boundaryOfBinaryTree(TreeNoderoot) {
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List<Integer>nodes =newArrayList<>(1000);
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if(root ==null)returnnodes;
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nodes.add(root.val);
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leftBoundary(root.left,nodes);
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leaves(root.left,nodes);
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leaves(root.right,nodes);
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rightBoundary(root.right,nodes);
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returnnodes;
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}
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publicvoidleftBoundary(TreeNoderoot,List<Integer>nodes) {
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if(root ==null || (root.left ==null &&root.right ==null))return;
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nodes.add(root.val);
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if(root.left ==null)leftBoundary(root.right,nodes);
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elseleftBoundary(root.left,nodes);
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}
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publicvoidrightBoundary(TreeNoderoot,List<Integer>nodes) {
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if(root ==null || (root.right ==null &&root.left ==null))return;
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if(root.right ==null)rightBoundary(root.left,nodes);
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elserightBoundary(root.right,nodes);
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nodes.add(root.val);// add after child visit(reverse)
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}
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publicvoidleaves(TreeNoderoot,List<Integer>nodes) {
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if(root ==null)return;
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if(root.left ==null &&root.right ==null) {
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nodes.add(root.val);
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return;
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}
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leaves(root.left,nodes);
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leaves(root.right,nodes);
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}
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//assume the tree has no duplicate values
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publicList<Integer>boundaryOfBinaryTree_failed_attempt(TreeNoderoot) {
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//this my original naive attemp failed by 79/117 test cases, specifically by {@link BoundaryofBinaryTreeTest.test5}.
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List<Integer>leftBoundary =newArrayList<>();
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leftBoundary =findLeftBoundary(root,leftBoundary);
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List<Integer>leaves =newArrayList<>();
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leaves =findLeaves(root,leaves);
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List<Integer>rightBoundary =newArrayList<>();
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rightBoundary =findRightBoundary(root,rightBoundary);
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List<Integer>boundary =newArrayList<>();
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Set<Integer>set =newHashSet<>();
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for (inti =0;i <leftBoundary.size();i++) {
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boundary.add(leftBoundary.get(i));
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set.add(leftBoundary.get(i));
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}
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for (inti =0;i <leaves.size();i++) {
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if (!set.add(leaves.get(i)))continue;
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boundary.add(leaves.get(i));
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}
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for (inti =rightBoundary.size()-1;i >=0;i--) {
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if (!set.add(rightBoundary.get(i)))continue;
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boundary.add(rightBoundary.get(i));
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}
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returnboundary;
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}
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privateList<Integer>findRightBoundary(TreeNoderoot,List<Integer>rightBoundary) {
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if (root ==null)returnrightBoundary;
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if (root.right !=null) {
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rightBoundary.add(root.right.val);
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findRightBoundary(root.right,rightBoundary);
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}elseif (root.left !=null) {
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rightBoundary.add(root.left.val);
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findRightBoundary(root.left,rightBoundary);
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}elseif (root.left ==null &&root.right ==null) {
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returnrightBoundary;
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}
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returnrightBoundary;
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}
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privateList<Integer>findLeaves(TreeNoderoot,List<Integer>leaves) {
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if (root ==null)returnleaves;
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if (root.left !=null) {
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findLeaves(root.left,leaves);
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}
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if (root.right !=null) {
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findLeaves(root.right,leaves);
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}
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if (root.left ==null &&root.right ==null) {
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leaves.add(root.val);
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}
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returnleaves;
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}
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privateList<Integer>findLeftBoundary(TreeNoderoot,List<Integer>leftBoundary) {
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if (root ==null) {
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returnleftBoundary;
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}
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if (root !=null) {
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leftBoundary.add(root.val);
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findLeftBoundary(root.left,leftBoundary);
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}
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returnleftBoundary;
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}
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}
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packagecom.stevesun;
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importcom.stevesun.common.classes.TreeNode;
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importcom.stevesun.solutions.BoundaryofBinaryTree;
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importorg.junit.Before;
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importorg.junit.BeforeClass;
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importorg.junit.Test;
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importjava.util.ArrayList;
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importjava.util.Arrays;
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importjava.util.List;
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importstaticjunit.framework.Assert.assertEquals;
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publicclassBoundaryofBinaryTreeTest {
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privatestaticBoundaryofBinaryTreetest;
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privatestaticList<Integer>expected;
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privatestaticList<Integer>actual;
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privatestaticTreeNoderoot;
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@BeforeClass
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publicstaticvoidsetup(){
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test =newBoundaryofBinaryTree();
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}
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@Before
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publicvoidsetupForEachTest(){}
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@Test
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publicvoidtest1(){
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root =newTreeNode(1);
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root.right =newTreeNode(2);
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root.right.left =newTreeNode(3);
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root.right.right =newTreeNode(4);
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actual =test.boundaryOfBinaryTree(root);
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expected =newArrayList<>(Arrays.asList(1,3,4 ,2));
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assertEquals(expected,actual);
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}
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@Test
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publicvoidtest2(){
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root =newTreeNode(1);
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root.left =newTreeNode(2);
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root.left.left =newTreeNode(4);
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root.left.right =newTreeNode(5);
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root.left.right.left =newTreeNode(7);
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root.left.right.right =newTreeNode(8);
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root.right =newTreeNode(3);
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root.right.left =newTreeNode(6);
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root.right.left.left =newTreeNode(9);
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root.right.left.right =newTreeNode(10);
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actual =test.boundaryOfBinaryTree(root);
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expected =newArrayList<>(Arrays.asList(1,2,4,7,8,9,10,6,3));
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assertEquals(expected,actual);
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}
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@Test
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publicvoidtest3(){
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root =newTreeNode(1);
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actual =test.boundaryOfBinaryTree(root);
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expected =newArrayList<>(Arrays.asList(1));
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assertEquals(expected,actual);
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}
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@Test
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publicvoidtest4(){
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root =newTreeNode(1);
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root.left =newTreeNode(2);
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root.left.left =newTreeNode(3);
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root.left.left.left=newTreeNode(4);
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actual =test.boundaryOfBinaryTree(root);
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expected =newArrayList<>(Arrays.asList(1,2,3,4));
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assertEquals(expected,actual);
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}
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@Test
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publicvoidtest5(){
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root =newTreeNode(1);
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root.left =newTreeNode(2);
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root.left.right =newTreeNode(4);
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root.right =newTreeNode(3);
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root.right.left =newTreeNode(5);
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root.left.right.left =newTreeNode(6);
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root.left.right.right =newTreeNode(7);
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actual =test.boundaryOfBinaryTree(root);
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expected =newArrayList<>(Arrays.asList(1,2,4,6,7,5,3));
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assertEquals(expected,actual);
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}
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}

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