题目:给定一颗树,判断其是否为平衡二叉树(平衡二叉树的左右子树高度差不大于1)
算法:递归遍历二叉树的左右子树高度,记录最大高度差
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
        public int maxDepthDiff = -1;
		
	    public boolean isBalanced(TreeNode root) {
	    	TreeDepth(root);
	        return (maxDepthDiff <= 1);
	    }
	    /**
	     * find the difference for left child depth and right child depth
	     * @param root
	     * @return
	     */
	    public int TreeDepth(TreeNode root) {
	    	if (null == root) {
	    		return 0;
	    	}
	    	
	    	int left = TreeDepth(root.left);
	    	int right = TreeDepth(root.right);
	    	
	    	int depthDiff = Math.abs(left-right);
	    	if (maxDepthDiff < depthDiff) {
	    		maxDepthDiff = depthDiff;
	    	}
	    	return (left > right ? left+1 : right+1);
	    }
}[LeetCode]Balanced Binary Tree,布布扣,bubuko.com
[LeetCode]Balanced Binary Tree
原文地址:http://blog.csdn.net/yeweiouyang/article/details/36650573