LeetCode Summary
  • Introduction
  • DP, 动态规划类
    • Best Time to Buy and Sell Stock III 题解
    • Best Time to Buy and Sell Stock 题解
    • Climbing Stairs 题解
    • Decode Ways 题解
    • Distinct Subsequences 题解
    • Edit Distance 题解
    • Interleaving String 题解
    • Longest Palindromic Substring 题解
    • Maximum Product Subarray 题解
    • Maximum Subarray 题解
    • Minimum Path Sum 题解
    • Palindrome Partitioning 题解
    • Palindrome Partitioning II 题解
    • Scramble String 题解
    • Triangle 题解
    • Unique Binary Search Trees 题解
    • Unique Paths II 题解
    • Word Break 题解
    • Word Break II 题解
  • list, 链表相关
    • Add Two Numbers 题解
    • Convert Sorted List to Binary Search Tree 题解
    • Copy List with Random Pointer 题解
    • Insertion Sort List 题解
    • LRU Cache 题解
    • Linked List Cycle 题解
    • Linked List Cycle II 题解
    • Merge Two Sorted Lists 题解
    • Merge k Sorted Lists 题解
    • Partition List 题解
    • Remove Duplicates from Sorted List 题解
    • Remove Duplicates from Sorted List II 题解
    • Remove Nth Node From End of List 题解
    • Reorder List 题解
    • Reverse Linked List II 题解
    • Reverse Nodes in k-Group 题解
    • Rotate List 题解
    • Sort List 题解
    • Swap Nodes in Pairs 题解
  • binary tree, 二叉树相关
    • Balanced Binary Tree 题解
    • Binary Tree Inorder Traversal 题解
    • Binary Tree Level Order Traversal 题解
    • Binary Tree Level Order Traversal II 题解
    • Binary Tree Maximum Path Sum 题解
    • Binary Tree Postorder Traversal 题解
    • Binary Tree Preorder Traversal 题解
    • Binary Tree Zigzag Level Order Traversal 题解
    • Construct Binary Tree from Inorder and Postorder Traversal 题解
    • Construct Binary Tree from Preorder and Inorder Traversal 题解
    • Convert Sorted List to Binary Search Tree 题解
    • Flatten Binary Tree to Linked List 题解
    • Maximum Depth of Binary Tree 题解
    • Minimum Depth of Binary Tree 题解
    • Path Sum 题解
    • Path Sum II 题解
    • Populating Next Right Pointers in Each Node 题解
    • Populating Next Right Pointers in Each Node II 题解
    • Recover Binary Search Tree 题解
    • Same Tree 题解
    • Sum Root to Leaf Numbers 题解
    • Symmetric Tree 题解
    • Unique Binary Search Trees 题解
    • Unique Binary Search Trees II 题解
    • Validate Binary Search Tree 题解
  • sort, 排序相关
    • 3Sum Closest 题解
    • 3Sum 题解
    • 4Sum 题解
    • Insert Interval 题解
    • Longest Consecutive Sequence 题解
    • Merge Intervals 题解
    • Merge Sorted Array 题解
    • Remove Duplicates from Sorted Array 题解
    • Remove Duplicates from Sorted Array II 题解
    • Sort Colors 题解
    • Two Sum 题解
  • search, 搜索相关
    • First Missing Positive 题解
    • Find Minimum in Rotated Sorted Array 题解
    • Find Minimum in Rotated Sorted Array II 题解
    • Median of Two Sorted Arrays 题解
    • Search Insert Position 题解
    • Search a 2D Matrix 题解
    • Search for a Range 题解
    • Search in Rotated Sorted Array 题解
    • Search in Rotated Sorted Array II 题解
    • Single Number 题解
    • Single Number II 题解
  • math, 数学类相关
    • Add Binary 题解
    • Add Two Numbers 题解
    • Divide Two Integers 题解
    • Gray Code 题解
    • Integer to Roman 题解
    • Multiply Strings 题解
    • Palindrome Number 题解
    • Plus One 题解
    • [Pow(x, n) 题解](math/Pow(x,-n).md)
    • Reverse Integer 题解
    • Roman to Integer 题解
    • [Sqrt(x) 题解](math/Sqrt(x).md)
    • [String to Integer (atoi) 题解](math/String-to-Integer-(atoi).md)
    • Valid Number 题解
  • string, 字符串处理相关
    • Anagrams 题解
    • Count and Say 题解
    • Evaluate Reverse Polish Notation 题解
    • [Implement strStr() 题解](string/Implement-strStr().md)
    • Length of Last Word 题解
    • Longest Common Prefix 题解
    • Longest Palindromic Substring 题解
    • Longest Substring Without Repeating Characters 题解
    • Longest Valid Parentheses 题解
    • Minimum Window Substring 题解
    • Regular Expression Matching 题解
    • Reverse Words in a String 题解
    • Simplify Path 题解
    • Text Justification 题解
    • Valid Parentheses 题解
    • Wildcard Matching 题解
    • ZigZag Conversion 题解
  • combination and permutation, 排列组合相关
    • Combinations 题解
    • Combination Sum 题解
    • Combination Sum II 题解
    • Letter Combinations of a Phone Number 题解
    • Next Permutation 题解
    • Palindrome Partitioning 题解
    • Permutation Sequence 题解
    • Permutations 题解
    • Permutations II 题解
    • Subsets 题解
    • Subsets II 题解
    • Unique Paths 题解
  • matrix, 二维数组, 矩阵相关
    • Rotate Image 题解
    • Set Matrix Zeroes 题解
    • Spiral Matrix 题解
    • Spiral Matrix II 题解
    • Maximal Rectangle 题解
  • 回溯, BFS/DFS
    • Clone Graph 题解
    • Generate Parentheses 题解
    • N-Queens 题解
    • N-Queens II 题解
    • Restore IP Addresses 题解
    • Sudoku Solver 题解
    • Surrounded Regions 题解
    • Word Ladder 题解
    • Word Ladder II 题解
    • Word Search 题解
  • greedy, 贪心
    • Best Time to Buy and Sell Stock II 题解
    • Jump Game 题解
    • Jump Game II 题解
  • 其他
    • Candy 题解
    • Container With Most Water 题解
    • Gas Station 题解
    • Gray Code 题解
    • Max Points on a Line 题解
    • [Pascal's Triangle 题解](others/Pascal's-Triangle.md)
    • [Pascal's Triangle II 题解](others/Pascal's-Triangle-II.md)
    • Remove Element 题解
    • Trapping Rain Water 题解
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  1. binary tree, 二叉树相关

Path Sum II 题解

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Last updated 5 years ago

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题目来源：

>

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,
              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1
return
[
   [5,4,11,2],
   [5,8,4,5]
]

解题思路：

跟思路一样，不过这题把路径存起来。

    void search(vector<int> &path, vector<vector<int> >&result, TreeNode* node, int target)
    {
        assert(node != NULL);
        int v = node->val;// Attention 0
        if(node->left == NULL && node->right == NULL)
        {
            if(target == v) 
            {
                path.push_back(v); //Attention 1
                result.push_back(path);
                path.pop_back();
            }
            return;
        }
        if(node->left)
        {
            path.push_back(v); //Attention 1
            search(path, result, node->left, target - v);
            path.pop_back();
        }
        if(node->right)
        {
            path.push_back(v); //Attention 1
            search(path, result, node->right, target - v);
            path.pop_back();
        }
    }
    vector<vector<int> > pathSum(TreeNode *root, int sum) 
    {
        vector<vector<int> > result;
        if(root == NULL) return result;
        vector<int> path;
        search(path, result, root, sum);
        return move(result);
    }

注意别被code中的表象所迷惑，将//Attention 1的代码提取到//Attention 0处。 path先后push_back会反映到递归调用里面去的。

不然应该下面这样写。

    void search2(vector<int> &path, vector<vector<int> >&result, TreeNode* node, int target)
    {
        if(node == NULL) return;//cannot check if target is 0 then putinto result, for this result has been checked before invoke the recursion
        int v = node->val;
        path.push_back(v);
        if(node->left == NULL && node->right == NULL)
        {
            if(target == v) 
                result.push_back(path);
        }
        search2(path, result, node->left, target - v);
        search2(path, result, node->right, target - v);
        path.pop_back();
    }
Path Sum II
Path Sum