【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；

 /**
  * 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)
             return false;

         int curSum = sum - root->val;

         if (!root->left && !root->right && curSum == )
             return true;

         return hasPathSum(root->left, curSum) || \
                 hasPathSum(root->right, curSum);

     }
 };

迭代的解法：

 class Solution {
 public:
     bool hasPathSum(TreeNode *root, int sum) {
         stack<TreeNode *> nodeStack;
         TreeNode *preNode = NULL;
         TreeNode *curNode = root;
         int curSum = ;

         while (curNode || !nodeStack.empty()) {
             while (curNode) {
                 nodeStack.push(curNode);
                 curSum += curNode->val;
                 curNode = curNode->left;
             }

             curNode = nodeStack.top();

             if (curNode->left == NULL && \
         　　　　　curNode->right == NULL && \
         　　　　　curSum == sum) {
                 return true;
             }

             if (curNode->right && preNode != curNode->right) {
                 curNode = curNode->right;
             } else {
                 preNode = curNode;
                 nodeStack.pop();
                 curSum -= curNode->val;
                 curNode = NULL;
             }
         }
         return false;
     }
 };

附录：

用迭代法遍历二叉树思路总结