剑指Offer:平衡二叉树Java/Python

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 $\red{前序遍历}$，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