leetCode(40):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.

搜索支路的方法

/**
 * 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 findPath(TreeNode* root, int nowSum,const int sum)
    {
    	nowSum += root->val;//当前值

        //假设已经是叶子结点，且和相等，则直接返回真
    	if (!root->left && !root->right && nowSum == sum)
    		return true;

    	bool found_left = false;
    	bool found_right = false;

    	if (root->left)	//左子树不为空。则在左子树中搜索
    		found_left = findPath(root->left, nowSum, sum);

    	if (!found_left && root->right)	//假设左子树没找到，则在右子树中搜索
    		found_right = findPath(root->right, nowSum, sum);	

    	return found_left || found_right;//仅仅要在一条支中上找到则返回真
    }

    bool hasPathSum(TreeNode* root, int sum) {
        if (root == NULL)
		    return false;
	    findPath(root, 0, sum);
    }
};