学步园

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

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

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

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]
]

思路：

递归post-order访问树的同时求和即可。这里采用了另外一个办法，与上述相似。

题解：

/**
 * Definition for binary tree
 * 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) {
        if (root == nullptr)
            return false;
        
        if (root->left == nullptr && root->right == nullptr)
        {
            if (sum == root->val)
                return true;
            else
                return false;
        }
        else
            return (root->left == nullptr ? 
                false : hasPathSum(root->left, sum - root->val)) ||
                (root->right == nullptr ?
                false : hasPathSum(root->right, sum - root->val));
            
    }
};

思路：

preorder访问树，迭代进行。用一个vector保存路径。

题解：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> scratch;
    vector<int> history;
    
    void traverse(TreeNode* node, int sum)
    {
        history.push_back(node->val);
        if (node->left == nullptr && node->right == nullptr && 
            accumulate(begin(history), end(history), 0) == sum)
            scratch.push_back(history);
        if (node->left != nullptr)
            traverse(node->left, sum);
        if (node->right != nullptr)
            traverse(node->right, sum);
        history.pop_back();
    }
    
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        scratch.clear();
        if (root == nullptr)
            return scratch;
        traverse(root, sum);
        return scratch;
    }
};