LeetCode Path Sum

最新推荐文章于 2020-07-07 13:07:47 发布

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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，该路径上的节点值之和等于给定的目标值。通过深度优先搜索(DFS)策略，递归地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.

题意：求树上是否存在一条路径使得和为sum。

思路：dfs到叶子节点的时候再判断和是否满足。

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