Path Sum

最新推荐文章于 2024-10-08 17:25:07 发布

原创最新推荐文章于 2024-10-08 17:25:07 发布 · 451 阅读

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本文介绍了一种算法，用于判断给定的二叉树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数值。通过递归方法实现了这一功能，并提供了具体的代码实现。

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.

Analysis:

1. Since the path must be root-to-leaf, root.left == null && root.right == null must be satisfied when sum is 0.

2. Since the value of each node can be either positive or negative, only sum == 0 should be checked.

3. RECURSIVE!!! RECURSIVE!!! RECURSIVE!!!

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) {      // empty tree
            return false;
        }
        
        sum -= root.val;
        if(sum == 0 && root.left == null && root.right == null) {          // find such a root-to-left path
            return true;
        }
        
        if(root.left != null && hasPathSum(root.left, sum)) {               // check the left subtree
            return true;
        }
        else if(root.right != null && hasPathSum(root.right, sum)) {        // check the right subtree
            return true;
        }
        else {
            return false;
        }
    }
}

Update 02/06/2014:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root==null) return false;
        if(root.left==null && root.right==null && sum-root.val==0) return true;     // leaf node 
        return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
    }
}