Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. Left boundary is defined as the path from root to the…

Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. Left boundary is defined as the path from root to the…

［抄题］: Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. Left boundary is defined as the path from root…

判断一棵树是否是平衡树,即左右子树的深度相差不超过1. 我们可以回顾下depth函数其实是Leetcode 104 Maximum Depth of Binary Tree 二叉树 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL…

Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes.  Left boundary is defined as the path from root to th…

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longestpath between any two nodes in a tree. This path may or may not pass through the root. Example:Given a binary tr…

Balanced Binary Tree [数据结构和算法]全面剖析树的各类遍历方法 描述 解析 递归分别判断每个节点的左右子树 该题是Easy的原因是该题可以很容易的想到时间复杂度为O(n^2)的方法.即按照定义,判断根节点左右子树的高度是不是相差1,递归判断左右子树是不是平衡的. 根据深度判断左右子树是否平衡 在计算树的高度的同时判断该树是不是平衡的. 即,先判断子树是不是平衡的,若是,则返回子树的高度:若不是,则返回一个非法的数字,如负数. 当一个节点是左右子树有一个不是平衡二叉树则不必继…

交换左右叶子节点 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void swapLR(TreeNode* root){ if(!root) retur…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…