1.题目描述

输入一棵二叉树，判断该二叉树是否是平衡二叉树。

2.算法描述

平衡二叉树是指每个结点的左右子树的高度之差不超过1的树。
利用二叉树的前序遍历\red{前序遍历}前序遍历，先判断根节点是否满足上述性质。然后判断左右子树是否满足上述性质。

3.代码描述

3.1.Java代码

public class Solution {
public boolean IsBalanced_Solution(TreeNode root) {
if(root == null)
return true;
int leftDepth = TreeDepth(root.left);
int rightDepth = TreeDepth(root.right);
if(leftDepth-rightDepth >= 2 || rightDepth-leftDepth >= 2)
return false;
return IsBalanced_Solution(root.left) && IsBalanced_Solution(root.right);
}
public int TreeDepth(TreeNode root) {
if(root == null)
return 0;
int leftDepth = TreeDepth(root.left);
int rightDepth = TreeDepth(root.right);
return Math.max(leftDepth, rightDepth) + 1;
}
}

3.2.Python代码

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def IsBalanced_Solution(self, pRoot):
        if not pRoot:
            return True
        leftDepth = self.TreeDepth(pRoot.left)
        rightDepth = self.TreeDepth(pRoot.right)
        if abs(leftDepth-rightDepth)>=2:
            return False
        return self.IsBalanced_Solution(pRoot.left) and self.IsBalanced_Solution(pRoot.right)
    def TreeDepth(self, pRoot):
        if not pRoot:
            return 0
        leftDepth = self.TreeDepth(pRoot.left)
        rightDepth = self.TreeDepth(pRoot.right)
        return max(leftDepth, rightDepth) + 1