Leetcode: Balanced Binary Tree

最新推荐文章于 2021-05-11 11:45:04 发布

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本文介绍了一种自顶向下的方法来判断给定的二叉树是否为高度平衡树。高度平衡树定义为每个节点的左右子树深度之差不超过1。文章还提供了一个从底部向上实现解决方案的示例。

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 ofevery node never differ by more than 1.

Top down solution.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isBalanced(TreeNode root) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if(root == null)
			return true;
		while(root != null){
			int left = maxHeight(root.left);
			int right = maxHeight(root.right);
			if(Math.abs(left - right) > 1)
				return false;
			else
				return isBalanced(root.left) && isBalanced(root.right);
		}
		return false;
	}
	
	public int maxHeight(TreeNode root){
		if(root == null)
			return 0;
		int left = maxHeight(root.left) + 1;
		int right = maxHeight(root.right) + 1;
		return left > right ? left : right;
	}
}

Bottom up solution.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isBalanced(TreeNode root){
		// Start typing your Java solution below
		// DO NOT write main() function
		return decide(root) >= 0 ? true : false;
	}
	public int decide(TreeNode root){
		if(root == null)
			return 0;
		int left = decide(root.left);
		int right = decide(root.right);
		if(left < 0 || right < 0 || Math.abs(left - right) > 1)
			return -1;
		return left > right ? left + 1 : right + 1;
	}
}