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.

解题思路:

由于需要找寻根到叶子节点的路径和，我们可以通过遍历一颗树即可得知，通常树的遍历有

四种方式:先序遍历、中序遍历、后序遍历、层次遍历，任选一种遍历方式即可.

解题代码(非递归):

/**
 * 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;
        queue
  
    > que;
        que.push(make_pair(root,root->val));
        while(!que.empty())
        {
            pair
   
     p = que.front(); que.pop(); if(p.first->left == NULL && p.first->right == NULL && p.second == sum) return true; if(p.first->left) que.push(make_pair(p.first->left,p.second + p.first->left->val)); if(p.first->right) que.push(make_pair(p.first->right,p.second + p.first->right->val)); } return false; } }; 
   
  

/**
 * 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 dfs(TreeNode *rt,long long sum)
    {
        if (rt->left == rt->right && !rt->left)
            return sum == rt->val ;
        if (rt->left && rt->right)
            return dfs(rt->left,sum - rt->val) | dfs(rt->right,sum - rt->val);
        return dfs(rt->left ? rt->left : rt->right, sum - rt->val);
    }
    bool hasPathSum(TreeNode *root, int sum) 
    {
        return root ? dfs(root,sum) : false ;   
    }
};