LeetCode 110 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.

思路1：使用自动向下的递归，每次递归都要计算左右子树的高度。若根节点是平衡的，则看其左右节点是否是平衡的，不断的递归。这样有重复。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
	public int maxDepth(TreeNode root){
		if(root==null) return 0;
		int maxleft=maxDepth(root.left);
		int maxright=maxDepth(root.right);
		return 1+(maxleft>maxright?maxleft:maxright);
	}
	public boolean isBalanced(TreeNode root) {
		if(root==null) return true;
		int leftdepth=maxDepth(root.left);
		int rightdepth=maxDepth(root.right);
		if(Math.abs(leftdepth-rightdepth)>1) return false;
		return isBalanced(root.left)&&isBalanced(root.right);
	}
}

思路2：如果我们用后序遍历的方式遍历二叉树的每一个结点，在遍历到一个结点之前我们已经遍历了它的左右子树。 只要在遍历每个结点的时候记录它的深度（某一结点的深度等于它到叶节点的路径的长度）， 我们就可以一边遍历一边判断每个结点是不是平衡的

由于Java无法像C那样“传递参数的地址，函数返回时能得到参数的值”，我们只能新建一个静态的内部类：Depth

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public static class  Depth{  
        private int depth;  
    }
    
    public boolean isBalanced(TreeNode root, Depth d) {
		if (root == null) {
			d.depth = 0;
			return true;
		}
		Depth left=new Depth();
		Depth right=new Depth();
		if (isBalanced(root.left, left) && isBalanced(root.right, right)) {
			int diff = left.depth - right.depth;
			if ((diff > 1) || (diff < -1)) {
				return false;
			}
			d.depth = 1 + (left.depth > right.depth ? left.depth : right.depth);
			return true;
		}
		return false;
	}
	
    public boolean isBalanced(TreeNode root) {
       	Depth temp=new Depth();
		return isBalanced(root,temp); 
    }
}