二叉树中每个节点能到达的最远距离

一、题目描述

输入一个只包含0和1的字符串表示二叉树，输出每个节点能到达的最远距离，通过父节点的路径也要考虑。

比如输入"111110000010000", 输出[3,2,4,3,3,4]。

          1A

        /       \

    1B        1C

    /  \         /  \

1D   1E    0   0

 / \    / \    / \   / \

0 0 0 1F 0 0 0 0

A->B->E->F为3， B->E->F为2，C->A->B->E->F为4，D->B->E->F或D->B->A->C为3，E->B->A->C为3，F->E->B->A->C为4，输出自然为[3,2,4,3,3,4]。

二、解题思路

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <unordered_map>
#include <queue>

using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

// Build tree using level order traversal
TreeNode* buildTree(const string& s) {
    if (s.empty() || s[0] == '0') return nullptr;
    queue<TreeNode*> q;
    TreeNode* root = new TreeNode(1);
    q.push(root);
    int index = 1;
    int currentVal = 2;
    while (!q.empty() && index < s.size()) {
        TreeNode* node = q.front();
        q.pop();
        // Left child
        if (index < s.size()) {
            if (s[index] == '1') {
                node->left = new TreeNode(currentVal++);
                q.push(node->left);
            }
            index++;
        }
        // Right child
        if (index < s.size()) {
            if (s[index] == '1') {
                node->right = new TreeNode(currentVal++);
                q.push(node->right);
            }
            index++;
        }
    }
    return root;
}

unordered_map<TreeNode*, pair<int, int>> dp;

pair<int, int> dfsDown(TreeNode* node) {
    if (!node) return {-1, -1}; // Base case: null node has depth -1
    auto left = dfsDown(node->left);
    auto right = dfsDown(node->right);
    int cl = left.first + 1; // Edge count to left child
    int cr = right.first + 1;
    int longest = max(cl, cr);
    int second;
    if (cl > cr) {
        second = max(left.second + 1, cr);
    } else if (cr > cl) {
        second = max(cl, right.second + 1);
    } else {
        second = max(left.second + 1, right.second + 1);
    }
    dp[node] = {longest, second};
    return {longest, second};
}

void dfsUp(TreeNode* node, TreeNode* parent, int up, vector<int>& result) {
    if (!node) return;
    int currentUp = 0;
    if (parent) {
        int siblingLongest = -1;
        if (parent->left == node && parent->right) {
            siblingLongest = dp[parent->right].first;
        } else if (parent->right == node && parent->left) {
            siblingLongest = dp[parent->left].first;
        }
        currentUp = max(up + 1, siblingLongest + 2);
    }
    result[node->val - 1] = max(dp[node].first, currentUp);
    dfsUp(node->left, node, currentUp, result);
    dfsUp(node->right, node, currentUp, result);
}

vector<int> maxDistances(string s) {
    TreeNode* root = buildTree(s);
    if (!root) return {};
    dp.clear();
    dfsDown(root);
    int maxNode = 0;
    for (auto& pair : dp) maxNode = max(maxNode, pair.first->val);
    vector<int> result(maxNode, 0);
    dfsUp(root, nullptr, 0, result);
    return result;
}

int main() {
    string input = "111110000010000";
    vector<int> output = maxDistances(input);
    for (int dist : output) {
        cout << dist << " ";
    }
    // Output: 3 2 4 3 3 4
    return 0;
}