【LeetCode & 剑指offer刷题】树题5：110 Balanced Binary Tree

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

110. 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.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

Return true.

 

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

Return false.

 

/**

 * Definition for a binary tree node.

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

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

 * };

 */

 

/*

对于每一个节点，我们通过checkDepth方法递归获得左右子树的深度，

如果子树是平衡的，则返回真实的深度，若不平衡，直接返回-1，

此方法时间复杂度O(N)，空间复杂度O(H)

 

对于用递归求解的题，可以画递归树来分析

*/

class Solution

{

public:   

    bool isBalanced(TreeNode *root)

    {

        if (checkDepth(root) == -1) return false;

        else return true;

    }

   

    //类似问题：104. Maximum Depth of Binary Tree

    int checkDepth(TreeNode *root)

    {

        if (!root) return 0; //为空时，返回深度0 

       

        int left = checkDepth(root->left); //返回左子树的深度，如果不平衡，返回-1

        if (left == -1) return -1; //只要哪里有-1一路返回，直到递归结束

       

        int right = checkDepth(root->right); //返回右子树的深度

        if (right == -1) return -1;

       

        int diff = abs(left - right); //计算左右子树的深度差，如果不平衡，返回-1，平衡则返回实际深度

        if (diff > 1)

            return -1;

        else

            return 1 + max(left, right);

    }

};