平衡二叉树

/*
问题描述：平衡二叉树，给定一个二叉树,确定它是高度平衡的。对于这个问题,一棵高度平衡的二叉树的定义是：一棵二叉树中每个节点的两个子树的深度相差不会超过1。
来源：LintCode
作者：syt

日期：2017-11-10

思路：计算每个结点的左子树和右子树的深度，并判断是否是平衡二叉树

*/

#include <iostream>class TreeNode{public:int val;TreeNode *left, *right;TreeNode(){}TreeNode(int val, TreeNode *lptr = NULL, TreeNode *rptr = NULL) {this->val = val;this->left = lptr;this->right = rptr;}};

#include "d_tree.h"#include <queue>using namespace std;/** @param root: The root of binary tree.* @return: True if this Binary tree is Balanced, or false.*/int calDepth(TreeNode* root){if (root == NULL)return 0;int left = calDepth(root->left);int right = calDepth(root->right);int max = 0;if (left > right)max = left + 1;elsemax = right + 1;return max;}bool isBalanced(TreeNode * root) {// write your code hereif (root == NULL)return true;queue<TreeNode *> q;TreeNode* p;q.push(root);while (!q.empty()){p = q.front();q.pop();int left = calDepth(p->left);int right = calDepth(p->right);int res = left - right;if (res > 1 || res < -1)return false;if (p->left != NULL)q.push(p->left);if (p->right != NULL)q.push(p->right);}return true;}TreeNode *buildTree(){TreeNode *root, *b, *c, *d, *e, *f, *g, *h;h = new TreeNode(8);g = new TreeNode(7);f = new TreeNode(6);e = new TreeNode(5);d = new TreeNode(4, g, h);c = new TreeNode(3, e, f);b = new TreeNode(2, d, (TreeNode *)NULL);root = new TreeNode(1, b, c);return root;}void main(){TreeNode *root = buildTree();cout << isBalanced(root) << endl;}