面试题34

输入一棵二叉树和一个整数，打印出二叉树中节点值的和为输入整数的所有路径。从树的根节点开始往下一直到叶节点所经过的节点形成一条路径。

示例:
给定如下二叉树，以及目标和 sum = 22，

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

返回:

[
   [5,4,11,2],
   [5,8,4,5]
]

提示：

• 节点总数 <= 10000

注意：本题与主站 113 题相同：https://leetcode-cn.com/problems/path-sum-ii/

Solutions

1. dfs with recursion

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void dfs(TreeNode * root, int sum) {
        if (!root) return;
        sum -= root->val;
        path.push_back(root->val);
        if (!root->left && !root->right) {
            if (sum == 0)
                res.push_back(path);
        }
        else {
            dfs(root->left, sum);
            dfs(root->right, sum);
        }
        path.pop_back();
    }
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        dfs(root, sum);
        return res;
    }
};

1. postorder traversal with stack

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        stack<TreeNode *> s;
        vector<vector<int>> res;
        vector<int> path;
        TreeNode * prev = nullptr;

        while (root || !s.empty()) {
            while (root) {
                s.push(root);
                sum -= root->val;
                path.push_back(root->val);
                root = root->left;
            }
            root = s.top();
            if (root->right && root->right != prev)
                root = root->right;
            else {
                if (!root->left && !root->right)
                    if (sum == 0)
                        res.push_back(path);
                path.pop_back();
                sum += root->val;
                prev = root; root = nullptr;
                s.pop();
            }
        }

        return res;
    }
};