Skip to content
On this page

110. 平衡二叉树

给定一个二叉树,判断它是否是高度平衡的二叉树。
本题中,一棵高度平衡二叉树定义为:

一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。

示例 1:

 3
/ \
9 20
  / \
 15 7

输入:root = [3,9,20,null,null,15,7]
输出:true

示例 2:

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

输入:root = [1,2,2,3,3,null,null,4,4]
输出:false

示例 3:
输入:root = []
输出:true

提示:
树中的节点数在范围 [0, 5000] 内
-104 <= Node.val <= 104

javascript
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isBalanced = function (root) {
  if (!root) return true
  if (Math.abs(maxDepth(root.left) - maxDepth(root.right)) > 1) return false
  return isBalanced(root.left) && isBalanced(root.right)
}

var maxDepth = function (root) {
  if (!root) return 0
  return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1
}

// test
function TreeNode(val, left, right) {
  this.val = val === undefined ? 0 : val
  this.left = left === undefined ? null : left
  this.right = right === undefined ? null : right
}

var n1 = new TreeNode(1)
var n2 = new TreeNode(2)
var n3 = new TreeNode(2)
var n4 = new TreeNode(3)
var n5 = new TreeNode(3)
var n6 = new TreeNode(4)
var n7 = new TreeNode(4)
n1.left = n2
n1.right = n3
n2.left = n4
n2.right = n5
n4.left = n6
n4.right = n7

console.log(isBalanced(n1))