LeetCode 110. Balanced Binary Tree

LeetCode 110. Balanced Binary Tree

Description:

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

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分析：

首先注意一下题意，这里需要判断每个结点的平衡性，不仅仅是根节点，一开始在这里搞错了，故需要递归判断左右子树的平衡性。

根节点为NULL，返回true，接着求左右子树的深度，可以利用104题的求左右子树的最大深度的代码，然后判断abs(left-right)<=1，即可得到结果。

代码如下：

#include <iostream>#include <cmath>using namespace std;/** * 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:    bool isBalanced(TreeNode* root) {        if (root == NULL) {            return true;        }        else {            int left = maxDepth(root->left);// 左子树深度            int right = maxDepth(root->right);// 右子树深度            // 注意这里依然需要递归左子树、右子树            return abs(left - right) <= 1 && isBalanced(root->left) && isBalanced(root->right);        }    }    int maxDepth(TreeNode* root) {        if (root == NULL) {            return 0;        }        else {            int leftDepth = maxDepth(root->left) + 1;            int rightDepth = maxDepth(root->right) + 1;            return leftDepth > rightDepth ? leftDepth : rightDepth;        }    }};// 构造二叉树int TreeNodeCreate(TreeNode* &tree) {    int val;    cin >> val;    if (val < 0) // 小于0表示空节点        tree = NULL;    else {        tree = new TreeNode(val); // 创建根节点        tree->val = val;        TreeNodeCreate(tree->left); // 创建左子树        TreeNodeCreate(tree->right);// 创建右子树    }    return 0;}int main() {    Solution s;    TreeNode* tree;    TreeNodeCreate(tree);    cout << s.isBalanced(tree) << endl;    return 0;}