leetcode

题目：Balanced Binary Tree

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.

个人思路：

1、判断每个节点（子树）的高度差，高度差在绝对值为1的范围内便是平衡二叉树

2、可以适当改造计算树高度的方法，即树的高度为左子树与右子树高度较大者加1

代码：

#include <stddef.h>
#include <iostream>
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution
{
public:
    bool isBalanced(TreeNode *root)
    {
        balanced = true;
        getDepth(root);

        return balanced;
    }
    int getDepth(TreeNode *root)
    {
        if (root == NULL)
        {
            return 0;
        }

        int leftDepth = getDepth(root->left);
        int rightDepth = getDepth(root->right);

        if (leftDepth - rightDepth > 1 || leftDepth - rightDepth < -1)
        {
            balanced = false;
        }

        return leftDepth > rightDepth ? leftDepth + 1 : rightDepth + 1;
    }
private:
    bool balanced;
};

int main()
{
    TreeNode *root = new TreeNode(1);
    root->right = new TreeNode(1);
    root->right->right = new TreeNode(1);
    Solution s;
    s.isBalanced(root);
    std::cout << root->val << std::endl << root->right->val << std::endl << root->right->val << std::endl;
    system("pause");

    return 0;
}

 上网搜了一些帖子，方法都是类似，就不贴出来了