判断平衡二叉树，只遍历一次的解法

LeetCode 110:

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]:

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

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

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

Return false.

采用后序遍历

/**

 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
static int x =[]()
{
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    return 0;
}();
class Solution {
public:
    bool isBalanced(TreeNode* root) {
        int depth = 0;
        return ifBalanced(root,&depth);
    }
private:
    bool ifBalanced(TreeNode* root,int* depth)
    {
        if(root==nullptr)
        {
            *depth = 0;
            return true;
        }
        int left,right;
        if(ifBalanced(root->left,&left)&&ifBalanced(root->right,&right))
        {
            if(abs(left-right)<=1)
            {
                *depth = left>right?left+1:right+1;
                return true;
            }
        }
        return false;
    }
};