Path Sum

最新推荐文章于 2024-10-08 17:25:07 发布

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#Algorithm #Interview #LeetCode #Tree #Depth-first Search

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本文探讨了如何确定一棵二叉树中是否存在从根节点到叶子节点的路径，使得路径上所有节点值之和等于给定的目标值。通过递归算法实现深度优先搜索，检查每个可能的路径。

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.

1st iteration:

/**
 * 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) {
        
        // invalid input
        if (root == NULL) {
            return false;
        }
        
        // leaf node
        if (root->left == NULL && root->right == NULL) {
            return root->val == sum;
        } 
        
        // depth-first-search
    
        // check on left child if it is not NULL
        if (root->left != NULL && hasPathSum(root->left, sum-root->val)) {
            return true;
        }
        
        // check on right child if it is not NULL
        if (root->right != NULL && hasPathSum(root->right, sum-root->val)) {
            return true;
        }
        
        return false;
    }
};

2nd iteration

/**
 * 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) {
       return hasPathSumR(root, 0, sum);
    }
    
    // traverse the tree dfs, add up sum when traversing, when
    // approach leaft, compare with target sum, if equal return
    // otherwise, give up the path
    bool hasPathSumR(TreeNode *root, int sumSoFar, int target) {
        if (root == NULL) {
            return false;
        }
        
        // add up sum when traversing
        sumSoFar += root->val;
        
        // when approach leaft, compare with target sum,  if equal return
        // otherwise, give up the path
        if (root->left == NULL && root->right == NULL) {
            return sumSoFar == target;
        }
        
        // traverse the tree dfs
        return hasPathSumR(root->left, sumSoFar, target) ||
               hasPathSumR(root->right, sumSoFar, target);
    }
};