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- 来自专栏calmound
Binary Tree Postorder Traversal
TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void PostOrder =NULL) { PostOrder(root->left,vec); PostOrder(root->right,vec); } } vector<int> postorderTraversal(TreeNode *root) { vector<int>vec; PostOrder
62180发布于 2018-04-17 - 来自专栏全栈程序员必看
Binary Tree Postorder Traversal — LeetCode
原题链接: http://oj.leetcode.com/problems/binary-tree-postorder-traversal/ 跟 Binary Tree Inorder Traversal
35710编辑于 2022-11-17 - 来自专栏搬砖记录
33 N-ary Tree Postorder Traversal
题目 Given an n-ary tree, return the postorder traversal of its nodes’ values. 再遍历根节点 解答 class Solution { // 存放结果 List<Integer> list = new ArrayList<>(); public List<Integer> postorder null) return list; // 递归遍历其他节点 for(Node node: root.children) postorder
42320发布于 2021-08-18 - 来自专栏皮皮星球
Binary Tree Postorder Traversal
Binary Tree Postorder Traversal Given a binary tree, return the postorder traversal of its nodes' values
54210发布于 2020-09-23 - 来自专栏计算机视觉与深度学习基础
Leetcode 145 Binary Tree Postorder Traversal
Given a binary tree, return the postorder traversal of its nodes' values.
51580发布于 2018-01-12 - 来自专栏Reck Zhang
LeetCode 0145 - Binary Tree Postorder Traversal
Binary Tree Postorder Traversal Desicription Given a binary tree, return the postorder traversal of its
33220发布于 2021-08-11 - 来自专栏赵俊的Java专栏
LeetCode 590 N-ary Tree Postorder Traversal
val = _val; children = _children; } }; */ class Solution { public List<Integer> postorder } }; */ class Solution { List<Integer> list = new ArrayList<>(); public List<Integer> postorder return list; } for (Node child : root.children) { postorder return list; } } Runtime: 1 ms, faster than 100.00% of Java online submissions for N-ary Tree Postorder Memory Usage: 48.2 MB, less than 40.66% of Java online submissions for N-ary Tree Postorder Traversal
57010发布于 2019-12-30 - 来自专栏SnailTyan
Leetcode145——Binary Tree Postorder Traversal
问题描述 Given a binary tree, return the postorder traversal of its nodes’ values.
40550发布于 2019-05-26 - 来自专栏尾尾部落
Construct Binary Tree from Inorder and Postorder Traversal
链接：https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder traversal of a , 0, postorder.length-1, inorderMap); } public TreeNode buildTree(int[] inorder, int iLeft, >iRight || pLeft>pRight) return null; TreeNode cur = new TreeNode(postorder[pRight , i+1, iRight, postorder, pLeft+i-iLeft, pRight-1, inorderMap); return cur; } }
62320发布于 2018-09-04 - 来自专栏乐行僧的博客
PAT（甲级）1138.Postorder Traversal（25）
PAT 1138.Postorder Traversal（25） Suppose that all the keys in a binary tree are distinct positive integers Given the preorder and inorder traversal sequences, you are supposed to output the first number of the postorder 输出格式： For each test case, print in one line the first number of the postorder traversal sequence of (node* root){ if(root){ postorder(root->lchild); postorder(root->rchild); pre[i]);} for(int i=1; i<=n; ++i){scanf("%d", &mid[i]);} node* root = build(1, n, 1, n); postorder
34030编辑于 2022-02-25 - 来自专栏给永远比拿愉快
Leetcode: Construct Binary Tree from Inorder and Postorder Traversal
题目： Given inorder and postorder traversal of a tree, construct the binary tree. ) { if (postorder.empty()) return nullptr; TreeNode *root = makeNode(inorder.begin (), inorder.end(), postorder.begin(), postorder.end()); return root; } }; Java参考代码： /** , postBegin + leftSize); root.right = makeNode(inorder, index + 1, inEnd, postorder, postBegin , postorder, 0, postorder.length); return root; } }
56620发布于 2019-01-22 - 来自专栏米扑专栏
【leetcode】Construct Binary Tree from Inorder and Postorder Traversal
Question： Given inorder and postorder traversal of a tree, construct the binary tree. [postorder[post_offset]]; root->right = buildTreeInOrder(postorder, post_offset-1, inorder, in_offset, in_offset-i); root->left = buildTreeInOrder(postorder, post_offset-(in_offset-i)-1 , postorder.size()-1, inorder, inorder.size()-1, postorder.size()); } }; 参考推荐： Construct Binary Tree from Inorder and Postorder Traversal
52530发布于 2019-02-19 - 来自专栏学习日记
Binary Tree Postorder Traversal.go
版权声明：原创勿转 https://blog.csdn.net/anakinsun/article/details/88992726
44130发布于 2019-04-12 - 来自专栏SnailTyan
N-ary Tree Postorder Traversal
val = _val; children = _children; } }; */ class Solution { public: vector<int> postorder val = _val; children = _children; } }; */ class Solution { public: vector<int> postorder reverse(result.begin(), result.end()); } }; Reference https://leetcode.com/problems/n-ary-tree-postorder-traversal
41820发布于 2019-05-25 - 来自专栏全栈程序员必看
Construct Binary Tree from Inorder and Postorder Traversal「建议收藏」
Given inorder and postorder traversal of a tree, construct the binary tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree ) { if (null == postorder || postorder.length < 1) { return null; } return buildTree(inorder, 0, inorder.length - 1, postorder, postorder.length - 1); } public TreeNode buildTree(int[] inorder, int start, int end, int[] postorder, int idx) {
26410编辑于 2022-08-04 - 来自专栏计算机视觉与深度学习基础
Leetcode 106 Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree. +1,ie); return root; } TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder ) { return dfs(postorder,0,inorder.size()-1,inorder,0,inorder.size()-1); } };
75560发布于 2018-01-12 - 来自专栏XINDOO的专栏
Leetcode Binary Tree Postorder Traversal(面试题推荐)
此题来自leetcode https://oj.leetcode.com/problems/binary-tree-postorder-traversal/ Given a binary tree , return the postorder traversal of its nodes' values.
42210发布于 2021-01-21 - 来自专栏Reck Zhang
LeetCode 0106 - Construct Binary Tree from Inorder and Postorder Traversal
Construct Binary Tree from Inorder and Postorder Traversal Desicription Given inorder and postorder traversal private: TreeNode* generateTree(int postStart, int postEnd, int inStart, int inEnd, vector<int>& postorder int inIndex = 0; for(int i = inStart; i <= inEnd; i++) { if(inorder[i] == postorder , inorder); root->right = generateTree(postEnd-inEnd+inIndex, postEnd-1, inIndex+1, inEnd, postorder ) { return generateTree(0, postorder.size()-1, 0, inorder.size()-1, postorder, inorder);
25230发布于 2021-08-11 - 来自专栏SnailTyan
Construct Binary Tree from Inorder and Postorder Traversal
NULL) {} * }; */ class Solution { public: TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder ) { int current = inorder.size() - 1; return build(postorder, inorder, current, 0, postorder.size () - 1); } private: TreeNode* build(vector<int>& postorder, vector<int>& inorder, int& current if(current == -1 || start > end) { return nullptr; } int value = postorder , inorder, current, index + 1, end); node->left = build(postorder, inorder, current, start, index
39710发布于 2019-05-25 - 来自专栏JNing的专栏
Construct Binary Tree from Inorder and Postorder Traversal
Problem Given inorder and postorder traversal of a tree, construct the binary tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree ): if not len(postorder): return None root = TreeNode(postorder[-1]) idx = inorder.index(root.val) root.left = self.buildTree(inorder[ : idx], postorder[ : idx]) root.right = self.buildTree(inorder[idx+1 : ], postorder[idx : -1]) return root
43030发布于 2019-02-25
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