543. Diameter of Binary Tree - Easy

最新推荐文章于 2025-07-29 14:24:32 发布

转载最新推荐文章于 2025-07-29 14:24:32 发布 · 94 阅读

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原文链接：http://www.cnblogs.com/fatttcat/p/10191046.html

本文探讨了计算二叉树直径的两种算法实现。首先介绍了基本的递归方法，时间复杂度为O(n^2)，随后提出了优化方案，将时间复杂度降低至O(n)，并详细解释了优化后的算法原理及代码实现。

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Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

Example:
Given a binary tree

Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

Note: The length of path between two nodes is represented by the number of edges between them.

time: O(n^2)　　-- worst case, space: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int diameterOfBinaryTree(TreeNode root) {
        if(root == null) {
            return 0;
        }
        int d = depth(root.left) + depth(root.right);
        int left = diameterOfBinaryTree(root.left);
        int right = diameterOfBinaryTree(root.right);
        
        return Math.max(d, Math.max(left, right));
    }
    
    public int depth(TreeNode node) {
        if(node == null) {
            return 0;
        }
        return 1 + Math.max(depth(node.left), depth(node.right));
    }
}

optimized:

time: O(n)　　-- post order traverse, visited each node once,

space: O(height)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int max = 1;
    
    public int diameterOfBinaryTree(TreeNode root) {
        maxDiamter(root);
        return max - 1;
    }
    
    public int maxDiamter(TreeNode node) {
        if (node == null) {
            return 0;
        }
        int left = maxDiamter(node.left);
        int right = maxDiamter(node.right);

        max = Math.max(max, left + right + 1);
        return 1 + Math.max(left, right);
    }
}

转载于:https://www.cnblogs.com/fatttcat/p/10191046.html