leetcode【112】Path Sum

问题描述：

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.

Note: A leaf is a node with no children.

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.

源码：

还是递归大法，时间69%，空间100%。

/**
 * 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 && root->val == sum)     return true;
        bool res = false;
        if(root->left)   res |= hasPathSum(root->left, sum - root->val);
        if(root->right)   res |= hasPathSum(root->right, sum - root->val);
        return res;
    }
};

再来看看非递归的。

/**
 * 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;
        queue<pair<TreeNode*, int>> st;
        st.push(make_pair(root, sum - root->val));
        while(!st.empty()){
            auto tmp = st.front();
            // cout<<tmp.second<<endl;
            st.pop();
            if(tmp.second==0 && !tmp.first->left && !tmp.first->right)   return true;
            // if(tmp.second<=0)   continue;
            if(tmp.first->left)  st.push(make_pair(tmp.first->left, tmp.second - tmp.first->left->val));
            if(tmp.first->right)  st.push(make_pair(tmp.first->right, tmp.second - tmp.first->right->val));
        }
        return false;
    }
};