Leetcode 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.

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解题思路：

递归地计算每个节点的左右子树深度，看其是否平衡。

计算左右子树深度可参考Maximum Depth of Binary Tree 的递归解法。

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 Java code:

/**
 * 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;
        }
        int diff = maxDepth(root.left) - maxDepth(root.right);
        if(diff > 1 || diff < -1 ) {
            return false;
        }
        return isBalanced(root.left) && isBalanced(root.right);
    }
    
    public int maxDepth(TreeNode root) {
        //use recursion
        if(root == null) {
            return 0;
        }
        int leftmax = maxDepth(root.left);
        int rightmax = maxDepth(root.right);
        return Math.max(leftmax, rightmax) + 1;   
    }
}

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20160607  time complexity: O(nlgn)

public class Solution {
    public boolean isBalanced(TreeNode root) {
        //time complexity: O(nlgn)
        //base case
        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 root) {  //time complexity: O(n)
        if (root == null) {
            return 0;
        }
        return Math.max(getHeight(root.left), getHeight(root.right)) + 1;
    }
}

Reference:

1. http://www.cnblogs.com/infinityu/archive/2013/05/11/3073411.html