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Leetcode[110] Balanced Binary Tree

08 Nov 2015

###Task1 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.

###Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        l = self.findHigh(root.left)
        r = self.findHigh(root.right)
        if abs(l - r) > 1:
            return False
        return self.isBalanced(root.left) and self.isBalanced(root.right)
    def findHigh(self, root):
        if not root:
            return 0
        l = self.findHigh(root.left)
        r = self.findHigh(root.right)
        return max(l, r) + 1

###Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    public int height(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        int left = height(root.left);
        int right = height(root.right);
        //ps: there is in fact three condition: left balance, right balance, and height difference less than 1
        if (left == -1 || right == -1 || Math.abs(left - right) > 1) {
            return -1;
        }
        return Math.max(left, right) + 1;
    }
    
    public boolean isBalanced(TreeNode root) {
        return height(root) != -1;
    }
}

###Points

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