leetcode543 - Diameter of Binary Tree - easy

最新推荐文章于 2025-07-29 14:24:32 发布

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最新推荐文章于 2025-07-29 14:24:32 发布

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原文链接：http://www.cnblogs.com/jasminemzy/p/9771834.html

本文深入探讨了计算二叉树直径的算法，即求解二叉树中最长路径的边数，无论路径是否通过根节点。通过递归方法，计算每个节点的左右子树深度，从而得出经过该节点的最长路径。最终，全局变量记录了所有节点的最长路径，返回其最大值作为结果。

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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

/ \

2 3

/ \

4 5

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.

树的递归

问题转化：对每个节点，计算左深度+右深度的值，就是经过该节点的最长path的edge数。

那如果对每个节点都得到过这个值，并且和全局变量打擂台，那就得到答案了呀。经过某个节点得到的最长path对应的长度就被存储在全局变量里了。

递归函数返回当前节点的深度给父亲用。递归函数内借助孩子打擂台。

实现：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int maxLen = 0;
    public int diameterOfBinaryTree(TreeNode root) {
        calHelper(root);
        return maxLen;
    }
    
    /*
    * return the current depth. leaf node has depth of 1.
    */
    private int calHelper(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftDepth = calHelper(root.left);
        int rightDepth = calHelper(root.right);
        maxLen = Math.max(maxLen, leftDepth + rightDepth);
        return Math.max(leftDepth, rightDepth) + 1;
    }
}