【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 == 0)
            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 = 0;
        
        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