【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