c++ 二叉树打印节点路径

leetcode上有道题是关于path sum的，我第一时间的做法是用二叉树的路径去与sum比较。所以需要去打印出二叉树的节点路径，以下是用回溯法。

#include<iostream>
#include<vector>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
    
};
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        vector<int> path;
        path.reserve(100);
        bool flag = false;
        printPaths(root, path, 0);
        return flag;
    }
    void printPaths(TreeNode* node, vector<int>& path, int pathlen) {
        if (node == NULL) return;
        // append this node to the path array 
        path[pathlen++] = node->val;
        // path.push_back(node->val);
        // it's a leaf, so print the path that led to here  
        if (node->left == NULL && node->right == NULL) {
            printArray(path, pathlen);
        }
        else {
            // otherwise try both subtrees  
            printPaths(node->left, path, pathlen);
            printPaths(node->right, path, pathlen);
            path.pop_back();
        }
    }
    void printArray(vector<int> &ints, int len) {
        for (int i = 0; i < len; i++) {
             cout << ints[i] << " " ; 
        }
         cout << endl;
    }
};

int main()
{
    Solution res;
    TreeNode t1(5);
    TreeNode t2(4);
    TreeNode t3(8);
    TreeNode t4(11);
    TreeNode t5(13);
    TreeNode t6(4);
    TreeNode t7(7);
    TreeNode t8(2);
    TreeNode t9(1);
    t1.left = &t2;
    t1.right = &t3;
    t2.left = &t4;
    t3.left = &t5;
    t3.right = &t6;
    t4.left = &t7;
    t4.right = &t8;
    t6.right = &t9;
    TreeNode * root = &t1;
    bool ret = res.hasPathSum(root,22);
    
    system("pause");
    return 0;
}

接着与sum做比较就能知道path sum是否相同了。

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        vector<int> path;
        path.reserve(100);
        bool flag = false;
        printPaths(root, path, 0, sum, flag);
        return flag;
    }
    void printPaths(TreeNode* node, vector<int>& path, int pathlen, int& sum, bool& flag) {
        if (node == NULL) return;
        // append this node to the path array 
        path[pathlen++] = node->val;
        // path.push_back(node->val);
        // it's a leaf, so print the path that led to here  
        if (node->left == NULL && node->right == NULL) {
            printArray(path, pathlen, sum, flag);
            if (flag) return;
        }
        else {
            // otherwise try both subtrees  
            printPaths(node->left, path, pathlen, sum, flag);
            printPaths(node->right, path, pathlen, sum, flag);
            path.pop_back();
        }
    }
    void printArray(vector<int> &ints, int len, int& sum, bool& flag) {
        int num = 0;
        for (int i = 0; i < len; i++) {
            // cout << ints[i] << " " ; 
            num += ints[i];
        }
        if (num == sum) flag = true;
        // cout << endl;
    }
};

不过这种方法不好，空间开销大，效率还不高。在网上看到这种解法贴一下地址：

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