leetCode 112. Path Sum 二叉樹問題

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.

思路：

使用遞歸先序遍歷。

代碼如下：

/**
 * 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(NULL == root)
            return false;
        return DFS(root,0,sum);
    }
    
    bool DFS(TreeNode * root,int curTotal,int sum)
    {
        if(NULL == root)
            return false;
        curTotal += root->val;
        if( !root->left && !root->right && (curTotal == sum))
            return true;
        else
            return DFS(root->left,curTotal,sum) || DFS(root->right,curTotal,sum);
    }
};

其他做法：

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

參考自：http://blog.csdn.net/booirror/article/details/42680111

2016-08-07 13:17:42