題目：

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.

解釋：
判斷一棵樹是不是平衡二叉樹，在dfs的過程中，如果子樹是平衡二叉樹，返回深度，否則返回-1。
python程式碼：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
 

#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        def TreeDepth(root):
            left_depth,right_depth=0,0
            if root.left:
 

                left_depth=TreeDepth(root.left)
            if root.right:
                right_depth=TreeDepth(root.right)
            if left_depth==-1 or right_depth==-1 or abs(left_depth-right_depth)>1:
                return -1
            return max(left_depth,right_depth)+1
        depth=0
        if root:
            depth=TreeDepth(root);
        return depth!=-1

c++程式碼：

/**
 * 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) {
        int depth=0;
        if (root)
            depth=TreeDepth(root);
        return depth!=-1;
    }
    int TreeDepth(TreeNode* root)
    {
        int left_depth=0,right_depth=0;
        if (root->left)
            left_depth=TreeDepth(root->left);
        if (root->right)
            right_depth=TreeDepth(root->right);
        if (left_depth==-1 || right_depth==-1 || abs(left_depth-right_depth)>1)
            return -1;
        return max(left_depth,right_depth)+1;
    }
};

總結：
如果子樹是平衡二叉樹，返回深度，否則返回-1。