LeetCode----path-sum

最新推荐文章于 2019-05-12 19:33:21 发布

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本文介绍了一种算法，用于判断二叉树中是否存在从根节点到叶子节点的路径，使得这条路径上所有节点值之和等于给定的数值。通过递归方式遍历二叉树，检查路径和是否匹配。

题目描述

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 andsum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.

----------------------------------------------------------------------------------------------------

/**
 * Definition for binary tree
 * 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 == nullptr)
            return false;
        if (root->val == sum && !root->left && !root->right)
            return true;
        sum = sum - root->val;
        return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
    }
};

您的代码已保存

/**

 *
 Definition for binary tree

 *
 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) {

        int

m = 0;

        return

preorder(root, sum, m);

    }

    bool
 preorder(TreeNode *root, int

sum, int

m){

        if

(root == nullptr || (m+root->val > sum && !root->left && !root->right))

            return

false;

        if

(m+root->val == sum && !root->left && !root->right)

            return

true;

        m
 = m + root->val;

        return

preorder(root->left, sum, m) || preorder(root->right, sum, m);

    }

};