110. Balanced Binary Tree

Question

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.

Thinking:

Method 1:递归

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

二刷

二刷的时候犯了一些错误，要对于每一个结点进行检测，不能只判断根结点。

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