Path Sum -- leetcode

标签：二叉树   traversal   path   leetcode   

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.

              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.

基本思路：

进行深度递归。

逐层从sum中扣去当前val值；到叶结点时，看剩下的sum和val值是否相等。

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

Path Sum -- leetcode

标签：二叉树   traversal   path   leetcode   

原文地址：http://blog.csdn.net/elton_xiao/article/details/45622193