110. Balanced Binary Tree

Posted 2020-08-08 Machelsky

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.

Solution 1:

思路：首先判断root.left和root.right是否有符合条件的depth，加入辅助函数，也就是104的maxdepth，然后再递归检查左子树和右子树。

/**
 * 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;
        }
        if(Math.abs(depth(root.left)-depth(root.right))<=1)
        {
            return isBalanced(root.left)&&isBalanced(root.right);
        }
        else
        {
            return false;
        }
    }
        
    public int depth(TreeNode root)
    {
        if(root==null)
        {
            return 0;
        }
        return Math.max(depth(root.left),depth(root.right))+1;
    }
        
    
}

Solution 2: 

discussion看到一种更巧妙的做法：

public boolean isBalanced(TreeNode root) {
    return getDepth(root) != -1;
}

private int getDepth(TreeNode root) {
    if (root != null) {
        int left = getDepth(root.left);
        int right = getDepth(root.right);
        
        return (left == -1 || right == -1 || Math.abs(left-right) > 1) ? -1 : Math.max(left, right) + 1;
    }
    
    return 0;
}

Define a getDepth method, which will return the height of the tree, if it is balanced, otherwise, return -1.

定义的这个getDepth行使两个功能，一是记录depth如果balanced，二是记录－1如果不balanced。

定义好一个recursion的函数很重要啊！！