"Leetcode" 112-path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such this 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           /   /           /  4         /  \              7    2      1

Return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Tags:tree Depth-first Search Solution 1:recursion

/** Definition for a binary tree node. * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * TreeNode (int x): Val (x), left (null), right (NULL) {} *}; */classSolution { Public:    BOOLHaspathsum (treenode* root,intsum) {        if(!root)return false; if(!root->left &&!root->Right ) {            if(root->val==sum)return true; Else                return false; }        returnHaspathsum (Root->left, sum-(root->val)) | | Haspathsum (Root->right, sum-(root->val)); }};

Solution 2:stack

BOOLHaspathsum (treenode* root,intsum) {    if(!root)return false; intCnt=root->Val; Stack<TreeNode*>Stk; Unordered_map<treenode*,BOOL>visited;    Stk.push (root); Visited[root]=true;  while(!Stk.empty ()) {TreeNode* top=Stk.top (); if(!top->left&&!top->Right ) {            if(cnt==sum)return true; }        if(top->left&&visited[top->left]==false) {Stk.push (top-Left ); Visited[top->left]=true; Cur+=top->left->Val; Continue; }        if(top->right&&visited[top->right]==false) {Stk.push (top-Right ); Visited[top->right]=true; Cur+=top->right->Val; Continue;        } stk.pop (); CNT-=top->Val; }    return false;}

"Leetcode" 112-path Sum