leetcode112. Path Sum

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本文探讨了二叉树中寻找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目标值的问题。提供了两种解决方案：一种是递归地检查每个叶子节点的路径和；另一种是通过深度优先搜索(DFS)递归减少目标值直至到达叶子节点。通过具体的示例和代码实现了这两种方法。

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112. 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.

解法一

找到叶子结点，判断累加和是否等于target。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int target;
    boolean ret;
    public boolean hasPathSum(TreeNode root, int sum) {
        target = sum;
        if (root == null) {
            return false;
        }
        helper(root, root.val);
        return ret;
    }
    public void helper(TreeNode root, int sum) {
        if (root.left == null && root.right == null) {
            if (sum == target) {
                ret = true;
                return;
            }
        }

        if (root.left != null)
            helper(root.left, sum + root.left.val);
        if (root.right != null)
            helper(root.right, sum + root.right.val);
    }
}

解法二

dfs，每次减去该结点的值，直到叶子结点，判断是否值为0。

/**
 * Definition for a binary tree node.
 * 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;
        }
        return hasPathSum(root.left, sum - root.val) 
        || hasPathSum(root.right, sum - root.val);
    }
}