【Leetcode】【Easy】Path Sum

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.

递归的解法：

只考虑当前结点的成败条件，对于儿子，则交给递归去做。

读到某个结点，计算路径val之和，之后判断，如果不是叶子结点，递归调用函数进入其子孙结点。如果是叶子结点，当val之和与sum值相等，返回true，不相等返回false。

注意：

1、注意题目是“root-to-leaf”即计算根节点到叶子节点的加和，不要只计算到某个枝干，即使运算到某个枝干时，sum值已经相等，也不能返回true；

2、注意可能出现正数负数混杂的情况；

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool hasPathSum(TreeNode *root, int sum) {
13         if (!root)
14             return false;
15             
16         int curSum = sum - root->val;
17         
18         if (!root->left && !root->right && curSum == 0)
19             return true;
20 
21         return hasPathSum(root->left, curSum) || 
22                 hasPathSum(root->right, curSum);
23         
24     }
25 };

迭代的解法：

 1 class Solution {
 2 public:
 3     bool hasPathSum(TreeNode *root, int sum) {
 4         stack<TreeNode *> nodeStack;
 5         TreeNode *preNode = NULL; 
 6         TreeNode *curNode = root;
 7         int curSum = 0;
 8         
 9         while (curNode || !nodeStack.empty()) {
10             while (curNode) {
11                 nodeStack.push(curNode);
12                 curSum += curNode->val;
13                 curNode = curNode->left;
14             }
15             
16             curNode = nodeStack.top();
17             
18             if (curNode->left == NULL && 
19         　　　　　curNode->right == NULL && 
20         　　　　　curSum == sum) {
21                 return true;
22             }
23             
24             if (curNode->right && preNode != curNode->right) {
25                 curNode = curNode->right;
26             } else {
27                 preNode = curNode;
28                 nodeStack.pop();
29                 curSum -= curNode->val;
30                 curNode = NULL;
31             }
32         }
33         return false;
34     }
35 };

附录：

用迭代法遍历二叉树思路总结