Path Sum II

描述

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:Given the below binary tree and sum = 22,

  1.           5
  2.          / \
  3.         4   8
  4.        /   / \
  5.       11  13  4
  6.      /  \    / \
  7.     7    2  5   1

return

  1. [
  2.    [5,4,11,2],
  3.    [5,8,4,5]
  4. ]

分析

跟上一题相比，本题是求路径本身。且要求出所有结果，左子树求到了满意结果，不能return，要接着求右子树。

代码

  1. // Path Sum II
  2. // 时间复杂度O(n)，空间复杂度O(logn)
  3. class Solution {
  4. public:
  5.     vector<vector<int> > pathSum(TreeNode *root, int sum) {
  6.         vector<vector<int> > result;
  7.         vector<int> cur; // 中间结果
  8.         pathSum(root, sum, cur, result);
  9.         return result;
  10.     }
  11. private:
  12.     void pathSum(TreeNode *root, int gap, vector<int> &cur,
  13.             vector<vector<int> > &result) {
  14.         if (root == nullptr) return;
  16.         cur.push_back(root->val);
  18.         if (root->left == nullptr && root->right == nullptr) { // leaf
  19.             if (gap == root->val)
  20.                 result.push_back(cur);
  21.         }
  22.         pathSum(root->left, gap - root->val, cur, result);
  23.         pathSum(root->right, gap - root->val, cur, result);
  25.         cur.pop_back();
  26.     }
  27. };

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原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/binary-tree/recursion/path-sum-ii.html