【leetcode】 Balanced Binary Tree

最新推荐文章于 2017-06-28 16:04:50 发布

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本文介绍了如何通过递归方式判断一个二叉树是否为高度平衡的二叉树，即左右子树深度之差不超过1。

From : https://leetcode.com/problems/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.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    
	bool isBalanced(TreeNode* root, int& dep) {
		if(!root) {dep=0; return true;}
		int left, right;
		if(isBalanced(root->left, left) && isBalanced(root->right, right)) {
			if(abs(right-left) <= 1) {
				dep = 1+max(left, right);
				return true;
			}
		}
		return false;
	}
    
    bool isBalanced(TreeNode *root) {
        int dep;
        return isBalanced(root, dep);
    }
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
	private static class Depth {
		int v = 0;
	}

	public boolean isBalanced(TreeNode root) {
		if (root == null) {
			return true;
		}
		return isBalanced(root, new Depth());
	}

	// dep is the deep of root
	boolean isBalanced(TreeNode root, Depth dep) {
		if (root == null) {
			dep.v = 0;
			return true;
		}
		Depth left = new Depth(), right = new Depth();
		if (isBalanced(root.left, left) && isBalanced(root.right, right)) {
			if (Math.abs(right.v - left.v) <= 1) {
				dep.v = 1 + Math.max(right.v, left.v);
				return true;
			}
		}
		return false;
	}
}