详细分析可参考LeetCode上的一篇博文。具体程序如下: 1vector<int> inorderTraversal(TreeNode*root) {2vector<int>rVec;3stack<TreeNode *>st;4TreeNode *tree =root;5while(tree || !st.empty())6{7if(tree)8{9st.push(tree);10tree = tree->left;11}12else13{14tree =st.top();15rVec.push...
classSolution {public: vector<int> preorderTraversal(TreeNode*root) {if(!root)return{}; vector<int>res; stack<TreeNode*>s{{root}};while(!s.empty()) { TreeNode*t =s.top(); s.pop(); res.push_back(t->val);if(t->right) s.push(t->right);if(t->left) s.push(t->left); ...
*/classSolution{public:vector<int>preorderTraversal(TreeNode*root){vector<int>ans;if(!root)returnans;stack<TreeNode*>s;s.push(root);while(!s.empty()){root=();s.pop();ans.push_back(root->val);if(root->right)s.push(root->right);if(root->left)s.push(root->left);}returnans;}}...
Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree. '#'representingnull"1,,3".Example 1: "9,3,4,#,#,1,#,#,2,#,6,#,#" ReturntrueExample 2: "1,#" Returnfalse...
Morris traversal: My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>ret=newArrayList<...
Given a binary tree, return the preorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 2 / 3 Output: [1,2,3] Follow up: Recursive solution is trivial, could you do it iteratively? 翻译 给出一棵二叉树,返回其节点值的前序遍历。
题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 根据前序遍历和中序遍历结果构造二叉树。 思路分析: 分析二叉树前序遍历和中序遍历的结果我们发现: 二叉树前序遍历的第一个节点是根节点。 在中序遍历...
105. Construct Binary Tree from Preorder and Inorder Traversal——tree,程序员大本营,技术文章内容聚合第一站。
https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/ """ # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): d...
来自专栏 · LeetCode刷题 Construct Binary Tree from Preorder and Inorder Traversal 题目描述(中等难度) 根据二叉树的先序遍历和中序遍历还原二叉树。 解法一 递归 先序遍历的顺序是根节点,左子树,右子树。中序遍历的顺序是左子树,根节点,右子树。