LeetCode 543. Diameter of Binary Tree

[woodyiiiiiii]

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

          1
         / \
        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.

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这是一道树形DP的Easy难度的题目，是求二叉树最长路径和的简化版，求二叉树最长路径长度。

class Solution {
    int diameter = 0;
    public int diameterOfBinaryTree(TreeNode root) {
        getDiameter(root, 1);
        return diameter;
    }
    public int getDiameter(TreeNode node, int l) {
        if (node == null) {
            return l - 1;
        }
        int left = getDiameter(node.left, l + 1);
        int right = getDiameter(node.right, l + 1);
        diameter = Math.max(diameter, left - l + right - l);
        return Math.max(left, right);
    }
}

需要求最长路径长度，就是要求两个节点的最大高度之差。
需要注意一点，当遍历到null时，返回的是当前高度减一(l - 1)，即当前根节点的高度，分析一下就能明白了。

大佬的做法更清晰，：

class Solution {
public:
    int diameterOfBinaryTree(TreeNode* root) {
        int res = 0;
        maxDepth(root, res);
        return res;
    }
    int maxDepth(TreeNode* node, int& res) {
        if (!node) return 0;
        int left = maxDepth(node->left, res);
        int right = maxDepth(node->right, res);
        res = max(res, left + right);
        return max(left, right) + 1;  // 最大路径+1，回溯
    }
};

从0开始，每次回溯都+1。
其实还可以用哈希表优化，记录每次遍历过的值，防止重复遍历：

class Solution {
public:
    int diameterOfBinaryTree(TreeNode* root) {
        int res = 0;
        maxDepth(root, res);
        return res;
    }
    int maxDepth(TreeNode* node, int& res) {
        if (!node) return 0;
        if (m.count(node)) return m[node];
        int left = maxDepth(node->left, res);
        int right = maxDepth(node->right, res);
        res = max(res, left + right);
        return m[node] = (max(left, right) + 1);
    }

private:
    unordered_map<TreeNode*, int> m;
};

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参考资料:
LeetCode原题