110. Balanced Binary Tree

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.

一刷
题解：
判断一个树是不是balanced tree. 首先要知道balanced tree的定义：一个树中的同一个level的子树的长度差不会超过1，所以只算root的左右子树的高度差是不够的。同样的，我们也可以思考怎么样让循环提前终止，当不满足要求时。

/**
 * 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 height(root)!=-1;
    }
    
    private int height(TreeNode root){
        if(root == null) return 0;
        int lh = height(root.left);
        if(lh == -1) return -1;
        int rh = height(root.right);
        if(rh == -1) return -1;
        if(Math.abs(lh-rh)>1) return -1;
        return Math.max(lh, rh)+1;
    }
}

二刷：
思路与一刷相同

/**
 * 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 height(root)!=-1;
    }
    
    private int height(TreeNode root){
        if(root == null) return 0;
        int lh = height(root.left);
        int rh = height(root.right);
        if(lh == -1 || rh == -1 || Math.abs(lh-rh)>1) return -1;
        return 1 + Math.max(lh, rh);
    }
}

三刷
利用求height的函数间接解，并且注意利用条件提前结束

/**
 * 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 height(root)!=-1;
    }
    
    public int height(TreeNode root){
        if(root == null) return 0;
        int left = height(root.left);
        int right = height(root.right);
        if(left == -1 || right == -1 || Math.abs(left-right)>1) return -1;
        return 1 + Math.max(left, right);
    }
}