题目

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.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

  1.     3
  2.    / \
  3.   9  20
  4.     /  \
  5.    15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

  1.        1
  2.       / \
  3.      2   2
  4.     / \
  5.    3   3
  6.   / \
  7.  4   4

Return false.

题意

判断指定树是否为平衡二叉树，即对于树中任意一个结点，它的左子树和右子树高度之差不大于1。

思路

在递归求各结点高度的同时，判断当前结点的左子树和右子树高度之差是否大于1。

代码实现

Java

  1. class Solution {
  2.     boolean isBalanced = true;
  4.     public boolean isBalanced(TreeNode root) {
  5.         getHeight(root);
  6.         return isBalanced;
  7.     }
  9.     private int getHeight(TreeNode x) {
  10.         // 空结点时返回高度
  11.         // 当已经确定该树不平衡时，无需再向下递归
  12.         if (x == null || !isBalanced) {
  13.             return 0;
  14.         }
  16.         int lHeight = getHeight(x.left);
  17.         int rHeight = getHeight(x.right);
  19.         if (Math.abs(lHeight - rHeight) > 1) {
  20.             isBalanced = false;
  21.         }
  23.         return Math.max(lHeight, rHeight) + 1;
  24.     }
  25. }

JavaScript

/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isBalanced = function (root) {
  return dfs(root) !== -1
}

let dfs = function (root) {
  if (!root) {
    return 0
  }

  let left = dfs(root.left)
  let right = dfs(root.right)

  if (left === -1 || right === -1 || Math.abs(left - right) > 1) {
    return -1
  }

  return Math.max(left, right) + 1
}