【LeetCode】112 - 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. 

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) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if(!root)return false;
        if(!root->left && !root->right ){
            if(root->val==sum)
                return true;
            else
                return false;
        }
        return hasPathSum(root->left, sum-(root->val)) || hasPathSum(root->right, sum-(root->val));
        
    }
};

 

 Solution 2:  stack

bool hasPathSum(TreeNode* root, int sum) {
    if(!root)return false;
    int cnt=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;
}