摘要：
Given a binary tree, determine if it is a valid binary search tree (BST).Assume a BST is defined as follows:The left subtree of a node contains only nodes with keys less than the node's key.The right subtree of a node contains only nodes with keys greater than the node's key.Both the left an 阅读全文
摘要：
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \3 4 4 3But the following is not: 1 / \ 2 2 \ \ 3 3 Note: Bonus points if you could solve it both recursively and itera... 阅读全文
摘要：
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7return its zigzag level order traversal as:[ [3], [20,9], [ 阅读全文
摘要：
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7return its zigzag level order traversal as:[ [3], [20,9], [ 阅读全文
摘要：
Given a binary tree, find its maximum depth.The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.class Solution {public: int maxDepth(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() f... 阅读全文
摘要：
Given preorder and inorder traversal of a tree, construct the binary tree.Note: You may assume that duplicates do not exist in the tree.class Solution {public: void f(vector &preorder,int s1, int e1,vector & inorder,int s2, int e2, TreeNode *& root){ if (s1 > e1){ return; ... 阅读全文
摘要：
Given inorder and postorder traversal of a tree, construct the binary tree.Note: You may assume that duplicates do not exist in the tree.class Solution {public: void f(vector &inorder,int s1, int e1,vector & postorder,int s2, int e2, TreeNode *& root){ if (s1 > e1){ return; ... 阅读全文