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); ...
Given a binary tree, return thepreordertraversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 1. 2. 3. 4. 5. return [1,2,3]. Note:Recursive solution is trivial, could you do it iteratively? 2.解决方案1 classS...
*/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;}}...
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。
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<...
在对节点TreeNode的抽象中,TreeNode具有三个属性,分别是val、left和right,val表示节点的值,而left和right分别指向左右两个子节点(默认为None)。 image.png 思路 需要一个长度可变的变量来存储结果。这里可以使用列表preorderlist。 从上面的分析,第一步是得到[A,A left,A right] ...
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...
971. Flip Binary Tree To Match Preorder Traversal # 题目 # You are given the root of a binary tree with n nodes, where each node is uniquely assigned a value from 1 to n. You are also given a sequence of n values voyage, which is the desired pre-order
LC 971. Flip Binary Tree To Match Preorder Traversal Given a binary tree with N nodes, each node has a different value from {1, ..., N}. A node in this b ... leetcode面试准备:Lowest Common Ancestor of a Binary Search Tree & Binary Tree leetcode面试准备:Lowest Common Ancestor ...