【Leetcode】一个recursive写的DFS 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.

Subscribe to see which companies asked this question

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if(root==null)
return false;
if(root.left==null && root.right==null)
return sum==root.val;
else{
int tmp = sum-root.val;
return hasPathSum(root.left,tmp) || hasPathSum(root.right,tmp);
}
}
}

再提供一个C++写的典型的DFS，拼命的先DFS到最左边的node，然后再慢慢stack去找有没有别的点

bool solution2(TreeNode *n, int sum){
if(!n) return false;
stack<TreeNode *> st;
TreeNode *cur = n, *pre;
while(cur || !st.empty()){
while(cur){
st.push(cur);
sum -= cur->val;
cur = cur->left;
}
cur = st.top();
if(!cur->left && !cur->right && !sum) return true;
if(cur->right && pre != cur->right) cur = cur->right;
else{
st.pop();
sum += cur->val;
pre = cur;
cur = NULL;
}
}
return false;
}