0110. Balanced Binary Tree (E)

Balanced Binary Tree (E)

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]:

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

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

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

Return false.

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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。

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代码实现

Java

class Solution {
    boolean isBalanced = true;

    public boolean isBalanced(TreeNode root) {
        getHeight(root);
        return isBalanced;
    }

    private int getHeight(TreeNode x) {
        // 空结点时返回高度
        // 当已经确定该树不平衡时，无需再向下递归
        if (x == null || !isBalanced) {
            return 0;
        }

        int lHeight = getHeight(x.left);
        int rHeight = getHeight(x.right);

        if (Math.abs(lHeight - rHeight) > 1) {
            isBalanced = false;
        }

        return Math.max(lHeight, rHeight) + 1;
    }
}

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
}