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  • leetCode(40):Path Sum 分类: leetCode 2015-07-17 16:46 125人阅读 评论(0) 收藏

    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
        
            //如果已经是叶子结点,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)	//如果左子树没找到,则在右子树中搜索
        		found_right = findPath(root->right, nowSum, sum);	
        
        	return found_left || found_right;//只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);
        }
    };



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  • 原文地址:https://www.cnblogs.com/zclzqbx/p/4687057.html
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