二叉树 得到路径

简介

以前觉得这个问题还是挺难的. 后来发现其实也很简单. 直接上代码.

code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include <iostream>
#include <vector>
using namespace std;
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x): val(x), left(nullptr), right(nullptr){}
};
    int level = 0;
    vector<TreeNode*> path1;
    vector<TreeNode*> path2;
    TreeNode* p;
    bool dfs(TreeNode *n, vector<TreeNode*> &path, TreeNode *p){
        if(n == p){
            path.push_back(n);
            return true;
        }
        if(n->left){
            if(dfs(n->left, path, p)){
                path.push_back(n);
                return true;
            }
        }
        if(n->right){
            if(dfs(n->right, path, p)){
                path.push_back(n);
                return true;
            }
        }
        return false;
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        
    }
int main(){
    TreeNode * a = new TreeNode(3);
    TreeNode * b = new TreeNode(5);
    TreeNode * c = new TreeNode(1);
    TreeNode * d = new TreeNode(6);
    TreeNode * e = new TreeNode(2);
    TreeNode * f = new TreeNode(0);
    TreeNode * g = new TreeNode(8);
    TreeNode * h = new TreeNode(7);
    TreeNode * i = new TreeNode(4);
    a->left = b;
    a->right = c;
    b->left = d;
    b->right = e;
    e->left = h;
    e->right = i;
    c->left = f;
    c->right = g;
    TreeNode * z = h;
    dfs(a, path1, z);
    for(auto it : path1){
        cout << it->val << std::endl;
    }
}