题目描述

题目难度：Easy
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.

AC代码1

仔细，认真

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    boolean isBalanced = true; //判断每个节点是否平衡
    public boolean isBalanced(TreeNode root) {
        if(root == null) return isBalanced;
        int[] heights = getHeight(root); 
        // 注意 & 的使用
        isBalanced &= Math.abs(heights[0] - heights[1]) <= 1 ? true : false;
        return isBalanced;
    }
    //求每个节点左右子树的高度
    private int[] getHeight(TreeNode root){
        int[] heights = new int[2];
        if(root == null) return heights;
        int[] leftHeights = getHeight(root.left);
        int[] rightHeights = getHeight(root.right);
        int leftMax = Math.max(leftHeights[0], leftHeights[1]);
        int rightMax = Math.max(rightHeights[0], rightHeights[1]);
        isBalanced &= Math.abs(leftMax - rightMax) <= 1 ? true : false;
        heights[0] = leftMax + 1;
        heights[1] = rightMax + 1;
        return heights;
    }
}

AC代码2

这个算法精辟的地方在于 check 函数返回值的含义：
如果当前节点保持平衡，则返回当前节点的高度；
否则返回-1,表示当前节点不是平衡的；

class Solution {
    public boolean isBalanced(TreeNode root) {
        if(root == null) return true;
     if(check(root) == -1) return false;
        return true;
    }
    
     public int check(TreeNode p){
        if(p == null) return 0;
        int left = check(p.left);
         if(left == -1) return -1;
        int right = check(p.right);
         if(right == -1) return -1;
        if(Math.abs(left - right) > 1) {
        	return -1;
        }
        return Math.max(left, right)+1;
    }

}