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简单

面试题 55 - II. 平衡二叉树

题目描述

输入一棵二叉树的根节点，判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。

 

示例 1:

给定二叉树 [3,9,20,null,null,15,7]

    3
   / \
  9  20
    /  \
   15   7

返回 true 。

示例 2:

给定二叉树 [1,2,2,3,3,null,null,4,4]

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

返回 false 。

 

限制：

  • 0 <= 树的结点个数 <= 10000

注意：本题与主站 110 题相同：https://leetcode.cn/problems/balanced-binary-tree/

 

解法

方法一：递归

我们设计一个递归函数 $dfs(root)$，函数返回值为 $root$ 节点的深度，如果 $root$ 节点不平衡，返回值为 $-1$。

函数 $dfs(root)$ 的递归过程如下：

  • 如果 $root$ 为空，返回 $0$。
  • 递归计算左右子树的深度，记为 $l$ 和 $r$。
  • 如果 $l$ 或 $r$ 为 $-1$，或者 $l$ 和 $r$ 的差的绝对值大于 $1$，返回 $-1$。
  • 否则，返回 $max(l, r) + 1$。

如果 $dfs(root)$ 返回值不为 $-1$，则说明 $root$ 节点平衡，返回 true，否则返回 false。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点数。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        def dfs(root):
            if root is None:
                return 0
            l, r = dfs(root.left), dfs(root.right)
            if l == -1 or r == -1 or abs(l - r) > 1:
                return -1
            return 1 + max(l, r)

        return dfs(root) != -1

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        return dfs(root) != -1;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int l = dfs(root.left);
        int r = dfs(root.right);
        if (l == -1 || r == -1 || Math.abs(l - r) > 1) {
            return -1;
        }
        return 1 + Math.max(l, r);
    }
}

C++

/**
 * 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) {
        function<int(TreeNode*)> dfs = [&](TreeNode* root) -> int {
            if (!root) {
                return 0;
            }
            int l = dfs(root->left);
            int r = dfs(root->right);
            if (l == -1 || r == -1 || abs(l - r) > 1) {
                return -1;
            }
            return 1 + max(l, r);
        };
        return dfs(root) != -1;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isBalanced(root *TreeNode) bool {
	var dfs func(*TreeNode) int
	dfs = func(root *TreeNode) int {
		if root == nil {
			return 0
		}
		l, r := dfs(root.Left), dfs(root.Right)
		if l == -1 || r == -1 || abs(l-r) > 1 {
			return -1
		}
		return 1 + max(l, r)
	}
	return dfs(root) != -1
}

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

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
        match root {
            None => 0,
            Some(node) => {
                let node = node.borrow();
                1 + Self::dfs(&node.left).max(Self::dfs(&node.right))
            }
        }
    }
    pub fn is_balanced(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
        match root {
            None => true,
            Some(node) => {
                let mut node = node.borrow_mut();
                let a = 10;
                (Self::dfs(&node.left) - Self::dfs(&node.right)).abs() <= 1
                    && Self::is_balanced(node.left.take())
                    && Self::is_balanced(node.right.take())
            }
        }
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isBalanced = function (root) {
    const dfs = root => {
        if (!root) {
            return 0;
        }
        const l = dfs(root.left);
        const r = dfs(root.right);
        if (l === -1 || r == -1 || Math.abs(l - r) > 1) {
            return -1;
        }
        return 1 + Math.max(l, r);
    };
    return dfs(root) !== -1;
};

C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public bool IsBalanced(TreeNode root) {
        return dfs(root) != -1;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int l = dfs(root.left);
        int r = dfs(root.right);
        if (l == -1 || r == -1 || Math.Abs(l - r) > 1) {
            return -1;
        }
        return 1 + Math.Max(l, r);
    }
}

Swift

/* public class TreeNode {
*     public var val: Int
*     public var left: TreeNode?
*     public var right: TreeNode?
*     public init(_ val: Int) {
*         self.val = val
*         self.left = nil
*         self.right = nil
*     }
* }
*/

class Solution {
    func isBalanced(_ root: TreeNode?) -> Bool {
        return dfs(root) != -1
    }

    private func dfs(_ root: TreeNode?) -> Int {
        guard let root = root else {
            return 0
        }
        let leftDepth = dfs(root.left)
        let rightDepth = dfs(root.right)
        if leftDepth == -1 || rightDepth == -1 || abs(leftDepth - rightDepth) > 1 {
            return -1
        }
        return 1 + max(leftDepth, rightDepth)
    }
}