110. Balanced Binary Tree - jiejackyzhang/leetcode-note GitHub Wiki

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.

Tree类题目。

一棵二叉树是height-balanced的条件是：

1. 它的左右子树高度差不超过1；
2. 它的左右子树也都是height-balanced的。

因此问题转换为一个recursive problem，关键点在于求出树的height。 求树的height也可以用recursion来解

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

上面的方法虽然比较直观，但是每次调用isBalanced方法都要将左右子树的height计算一遍，做了很多重复的工作。 可以只遍历一遍tree，计算height的同时返回是否balanced。 本质上是post-order traversal。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isBalanced(TreeNode root) {
        return root == null || helper(root) != -1;
    }
    
    private int helper(TreeNode node) {
        if(node.left == null && node.right == null) return 1;
        int lh = 0, rh = 0;
        if(node.left != null) lh = helper(node.left);
        if(node.right != null) rh = helper(node.right);
        if(lh == -1 || rh == -1) return -1;
        if(Math.abs(lh - rh) > 1) return -1;
        return Math.max(lh, rh) + 1;
    }
}