【小熊刷题】Balanced Binary Tree

最新推荐文章于 2020-03-07 22:18:27 发布

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最新推荐文章于 2020-03-07 22:18:27 发布

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本文深入探讨了如何通过递归算法判断并优化平衡二叉树，介绍了两种方法：传统的上行遍历和改进的下行遍历。详细解释了每个方法的实现逻辑、时间和空间复杂度，并对比了它们的效率。此外，文章还提供了一个实际的代码示例，帮助读者更好地理解并应用这些算法。

Question

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 differs by more than 1.

https://oj.leetcode.com/problems/balanced-binary-tree/

*Difficulty: Easy, Frequency: High

My Solution - Top-down recursion

runtime: O(n^2), stack space O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isBalanced(TreeNode root) {
        if(root == null) return true;
        if(Math.abs(depth(root.left)-depth(root.right)) > 1) return false;
        return isBalanced(root.left) && isBalanced(root.right);
    }

    public int depth(TreeNode node){
        int depth = 0;
        if(node == null) return 0;
        return Math.max(depth(node.left),depth(node.right))+1;
    }
}

Bottom-up recursion

上面的方法，会对每一个node的maxDepth进行重复计算，故用bottom-up的方法更有效。从leaf往上，如果左右有不是balanced记depth为-1，看到-1立即返回

O(n) runtime, O(n) stack space

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isBalanced(TreeNode root) {
        if(root == null) return true;
        if(Math.abs(depth(root.left)-depth(root.right)) > 1) return false;
        return isBalanced(root.left) && isBalanced(root.right);
    }

    public int depth(TreeNode node){
        int depth = 0;
        if(node == null) return 0;
        return Math.max(depth(node.left),depth(node.right))+1;
    }
}