letecode [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.

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.

 

题目大意：

　　给定二叉树，判断是否存在某条路径从根节点到叶节点，所有节点元素的和等于给定值。

理解：

　　遍历二叉树，每遍历到叶节点则判断是否等于给定值。

　　遍历时，给定值动态变化，每遍历某个节点，递归遍历它的子节点时，用总和-该节点值。

　　直接用==结果作为返回值。

代码C++：

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

运行结果：

　　执行用时 : 16 ms　　内存消耗 : 19.9 MB