|
| 1 | +/** |
| 2 | + * We are given a binary tree (with root node root), a target node, and an |
| 3 | + * integer value K. |
| 4 | + * |
| 5 | + * Return a list of the values of all nodes that have a distance K from the |
| 6 | + * target node. The answer can be returned in any order. |
| 7 | + * |
| 8 | + * Example 1: |
| 9 | + * Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2 |
| 10 | + * Output: [7,4,1] |
| 11 | + * |
| 12 | + * Explanation: |
| 13 | + * The nodes that are a distance 2 from the target node (with value 5) |
| 14 | + * have values 7, 4, and 1. |
| 15 | + * |
| 16 | + * https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/28/sketch0.png |
| 17 | + * |
| 18 | + * Note that the inputs "root" and "target" are actually TreeNodes. |
| 19 | + * The descriptions of the inputs above are just serializations of these objects. |
| 20 | + * |
| 21 | + * Note: |
| 22 | + * The given tree is non-empty. |
| 23 | + * Each node in the tree has unique values 0 <= node.val <= 500. |
| 24 | + * The target node is a node in the tree. |
| 25 | + * 0 <= K <= 1000. |
| 26 | + */ |
| 27 | + |
| 28 | +/** |
| 29 | + * Definition for a binary tree node. |
| 30 | + * public class TreeNode { |
| 31 | + * int val; |
| 32 | + * TreeNode left; |
| 33 | + * TreeNode right; |
| 34 | + * TreeNode(int x) { val = x; } |
| 35 | + * } |
| 36 | + */ |
| 37 | + |
| 38 | +publicclassAllNodesDistanceKInBinaryTree863 { |
| 39 | +publicList<Integer>distanceK(TreeNoderoot,TreeNodetarget,intK) { |
| 40 | +List<Integer>res =newArrayList<>(); |
| 41 | +if (root ==null)returnres; |
| 42 | +if (K ==0) { |
| 43 | +res.add(target.val); |
| 44 | +returnres; |
| 45 | + } |
| 46 | +distanceToRoot(root,target,K,res,0); |
| 47 | +returnres; |
| 48 | + } |
| 49 | + |
| 50 | +privateintdistanceToRoot(TreeNoderoot,TreeNodetarget,intK,List<Integer>res,intlevel) { |
| 51 | +if (root ==null)return -1; |
| 52 | +if (root.val ==target.val) { |
| 53 | +distanceKChildren(root,K,res); |
| 54 | +returnlevel; |
| 55 | + } |
| 56 | +intleftFlag =distanceToRoot(root.left,target,K,res,level +1); |
| 57 | +if (leftFlag != -1) { |
| 58 | +if (leftFlag -level ==K) { |
| 59 | +res.add(root.val); |
| 60 | + }elseif (leftFlag -level <K) { |
| 61 | +distanceKChildren(root.right,K - (leftFlag -level) -1,res); |
| 62 | + } |
| 63 | + } |
| 64 | +intrightFlag =distanceToRoot(root.right,target,K,res,level +1); |
| 65 | +if (rightFlag != -1) { |
| 66 | +if (rightFlag -level ==K) { |
| 67 | +res.add(root.val); |
| 68 | + }elseif (rightFlag -level <K) { |
| 69 | +distanceKChildren(root.left,K - (rightFlag -level) -1,res); |
| 70 | + } |
| 71 | + } |
| 72 | +returnleftFlag != -1 ?leftFlag :rightFlag; |
| 73 | + } |
| 74 | + |
| 75 | +privatevoiddistanceKChildren(TreeNoderoot,intK,List<Integer>res) { |
| 76 | +if (root ==null)return; |
| 77 | +Queue<TreeNode>q =newLinkedList<>(); |
| 78 | +q.add(root); |
| 79 | +inti =0; |
| 80 | +while (!q.isEmpty() &&i <K) { |
| 81 | +intsize =q.size(); |
| 82 | +for (intj=0;j<size;j++) { |
| 83 | +TreeNodecurr =q.poll(); |
| 84 | +if (curr.left !=null)q.add(curr.left); |
| 85 | +if (curr.right !=null)q.add(curr.right); |
| 86 | + } |
| 87 | +i++; |
| 88 | + } |
| 89 | +while (!q.isEmpty()) { |
| 90 | +res.add(q.poll().val); |
| 91 | + } |
| 92 | + } |
| 93 | + |
| 94 | + |
| 95 | +/** |
| 96 | + * https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/solution/ |
| 97 | + */ |
| 98 | +Map<TreeNode,TreeNode>parent; |
| 99 | +publicList<Integer>distanceK2(TreeNoderoot,TreeNodetarget,intK) { |
| 100 | +parent =newHashMap(); |
| 101 | +dfs(root,null); |
| 102 | + |
| 103 | +Queue<TreeNode>queue =newLinkedList(); |
| 104 | +queue.add(null); |
| 105 | +queue.add(target); |
| 106 | + |
| 107 | +Set<TreeNode>seen =newHashSet(); |
| 108 | +seen.add(target); |
| 109 | +seen.add(null); |
| 110 | + |
| 111 | +intdist =0; |
| 112 | +while (!queue.isEmpty()) { |
| 113 | +TreeNodenode =queue.poll(); |
| 114 | +if (node ==null) { |
| 115 | +if (dist ==K) { |
| 116 | +List<Integer>ans =newArrayList(); |
| 117 | +for (TreeNoden:queue) |
| 118 | +ans.add(n.val); |
| 119 | +returnans; |
| 120 | + } |
| 121 | +queue.offer(null); |
| 122 | +dist++; |
| 123 | + }else { |
| 124 | +if (!seen.contains(node.left)) { |
| 125 | +seen.add(node.left); |
| 126 | +queue.offer(node.left); |
| 127 | + } |
| 128 | +if (!seen.contains(node.right)) { |
| 129 | +seen.add(node.right); |
| 130 | +queue.offer(node.right); |
| 131 | + } |
| 132 | +TreeNodepar =parent.get(node); |
| 133 | +if (!seen.contains(par)) { |
| 134 | +seen.add(par); |
| 135 | +queue.offer(par); |
| 136 | + } |
| 137 | + } |
| 138 | + } |
| 139 | + |
| 140 | +returnnewArrayList<Integer>(); |
| 141 | + } |
| 142 | + |
| 143 | +publicvoiddfs(TreeNodenode,TreeNodepar) { |
| 144 | +if (node !=null) { |
| 145 | +parent.put(node,par); |
| 146 | +dfs(node.left,node); |
| 147 | +dfs(node.right,node); |
| 148 | + } |
| 149 | + } |
| 150 | + |
| 151 | + |
| 152 | +/** |
| 153 | + * https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/solution/ |
| 154 | + */ |
| 155 | +List<Integer>ans; |
| 156 | +TreeNodetarget; |
| 157 | +intK; |
| 158 | +publicList<Integer>distanceK3(TreeNoderoot,TreeNodetarget,intK) { |
| 159 | +ans =newLinkedList(); |
| 160 | +this.target =target; |
| 161 | +this.K =K; |
| 162 | +dfs(root); |
| 163 | +returnans; |
| 164 | + } |
| 165 | + |
| 166 | +// Return distance from node to target if exists, else -1 |
| 167 | +publicintdfs(TreeNodenode) { |
| 168 | +if (node ==null) |
| 169 | +return -1; |
| 170 | +elseif (node ==target) { |
| 171 | +subtree_add(node,0); |
| 172 | +return1; |
| 173 | + }else { |
| 174 | +intL =dfs(node.left),R =dfs(node.right); |
| 175 | +if (L != -1) { |
| 176 | +if (L ==K)ans.add(node.val); |
| 177 | +subtree_add(node.right,L +1); |
| 178 | +returnL +1; |
| 179 | + }elseif (R != -1) { |
| 180 | +if (R ==K)ans.add(node.val); |
| 181 | +subtree_add(node.left,R +1); |
| 182 | +returnR +1; |
| 183 | + }else { |
| 184 | +return -1; |
| 185 | + } |
| 186 | + } |
| 187 | + } |
| 188 | + |
| 189 | +// Add all nodes 'K - dist' from the node to answer. |
| 190 | +publicvoidsubtree_add(TreeNodenode,intdist) { |
| 191 | +if (node ==null)return; |
| 192 | +if (dist ==K) |
| 193 | +ans.add(node.val); |
| 194 | +else { |
| 195 | +subtree_add(node.left,dist +1); |
| 196 | +subtree_add(node.right,dist +1); |
| 197 | + } |
| 198 | + } |
| 199 | + |
| 200 | + |
| 201 | +Map<TreeNode,Integer>map =newHashMap<>(); |
| 202 | +publicList<Integer>distanceK4(TreeNoderoot,TreeNodetarget,intK) { |
| 203 | +List<Integer>res =newLinkedList<>(); |
| 204 | +find(root,target,K); |
| 205 | +dfs(root,target,K,map.get(root),res); |
| 206 | +returnres; |
| 207 | + } |
| 208 | + |
| 209 | +// find target node first and store the distance in that path that we could use it later directly |
| 210 | +privateintfind(TreeNoderoot,TreeNodetarget,intK) { |
| 211 | +if (root ==null)return -1; |
| 212 | +if (root ==target) { |
| 213 | +map.put(root,0); |
| 214 | +return0; |
| 215 | + } |
| 216 | +intleft =find(root.left,target,K); |
| 217 | +if (left >=0) { |
| 218 | +map.put(root,left +1); |
| 219 | +returnleft +1; |
| 220 | + } |
| 221 | +intright =find(root.right,target,K); |
| 222 | +if (right >=0) { |
| 223 | +map.put(root,right +1); |
| 224 | +returnright +1; |
| 225 | + } |
| 226 | +return -1; |
| 227 | + } |
| 228 | + |
| 229 | +privatevoiddfs(TreeNoderoot,TreeNodetarget,intK,intlength,List<Integer>res) { |
| 230 | +if (root ==null)return; |
| 231 | +if (map.containsKey(root))length =map.get(root); |
| 232 | +if (length ==K)res.add(root.val); |
| 233 | +dfs(root.left,target,K,length +1,res); |
| 234 | +dfs(root.right,target,K,length +1,res); |
| 235 | + } |
| 236 | + |
| 237 | +} |