leetcode-110-Balanced-Binary-Tree

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描述

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.

分析

判断一棵树是不是平衡二叉树。平衡二叉树的定义是：它是空树，或者，左右两个子树的高度差不超过1而且两个子树都是平衡二叉树。按照原理写代码就好写了，如果根节点是null，自然是二叉树，否则，需要满足两个条件，两子树高度差不超过1，并且两子树都是平衡二叉树。

解决方案1(C++)

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/**
 * 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:
    int depth(TreeNode* root) {
        return root?max(depth(root->left), depth(root->right))+1:0;
    }

    bool isBalanced(TreeNode* root) {
        return root?abs(depth(root->left)-depth(root->right))<=1 && isBalanced(root->left) && isBalanced(root->right) : true;
    }
};

解决方案2(Java)

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int depth(TreeNode root) {
        if(root == null) {
            return 0;
        }
        return Math.max(depth(root.left), depth(root.right)) + 1;
    }
    
    public boolean isBalanced(TreeNode root) {
        if(root == null) {
            return true;
        }
        return Math.abs(depth(root.left)-depth(root.right))<=1 && isBalanced(root.left) && isBalanced(root.right);
    }
}

(OS:果然还是C++简洁一点）

解决方案3(Python)

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# 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
        """
        if root is None:
            return True
        return abs(self.depth(root.left)-self.depth(root.right)) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right)
    
    def depth(self, root):
        if root is None:
            return 0
        return max(self.depth(root.left), self.depth(root.right))+1

解决方案4（Golang）

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isBalanced(root *TreeNode) bool {
    if root == nil {
        return true
    }
    return abs(height(root.Left)-height(root.Right)) <= 1 && isBalanced(root.Left) && isBalanced(root.Right)
}

func height(root *TreeNode) int {
    if root == nil {
        return 0
    }
    return max(height(root.Left), height(root.Right)) + 1
    
}

func max(x, y int) int {
    if x > y {
        return x
    }
    return y
}

func abs(x int) int {
    if x < 0 {
        return -1 * x
    }
    return x
}

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