Path-sum

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于 2018-07-24 16:00:50 发布

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本文链接：https://blog.youkuaiyun.com/Neo233/article/details/81186383

本文探讨了如何在二叉树中寻找一条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，详细解释了判断条件和递归终止条件。

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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
 * 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(sum - root.val == 0 && root.right == null && root.left == null)
            return true;
        return hasPathSum(root.right,sum-root.val)||(hasPathSum(root.left,sum-root.val));
    }
}