递归 二叉树

https://leetcode-cn.com/problems/path-sum/

 

给你二叉树的根节点 root 和一个表示目标和的整数 targetSum 。判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。如果存在，返回 true ；否则，返回 false 。

叶子节点 是指没有子节点的节点。

 

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
输出：true
解释：等于目标和的根节点到叶节点路径如上图所示。

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：false
解释：树中存在两条根节点到叶子节点的路径：
(1 --> 2): 和为 3
(1 --> 3): 和为 4
不存在 sum = 5 的根节点到叶子节点的路径。

示例 3：

输入：root = [], targetSum = 0
输出：false
解释：由于树是空的，所以不存在根节点到叶子节点的路径。

 

提示：

• 树中节点的数目在范围 [0, 5000] 内
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

112. Path Sum

难度简单858

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

 

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.

Example 2:

Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There two root-to-leaf paths in the tree:
(1 --> 2): The sum is 3.
(1 --> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.

Example 3:

Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.

 

Constraints:

• The number of nodes in the tree is in the range [0, 5000].
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

 

 

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func hasPathSum(root *TreeNode, targetSum int) bool {
	if root == nil {
		return false
	}
	if root.Val == targetSum {
		if root.Left == nil && root.Right == nil {
			return true
		}
	} else {
		if root.Left == nil && root.Right == nil {
			return false
		}
	}
	return hasPathSum(root.Left, targetSum-root.Val) || hasPathSum(root.Right, targetSum-root.Val)
}

　　

func hasPathSum(root *TreeNode, sum int) bool {
    if root == nil {
        return false
    }
    if root.Left == nil && root.Right == nil {
        return sum == root.Val
    }
    return hasPathSum(root.Left, sum - root.Val) || hasPathSum(root.Right, sum - root.Val)
} 

　　https://leetcode-cn.com/problems/path-sum/solution/lu-jing-zong-he-by-leetcode-solution/

 

标签：return,递归,sum,二叉树,false,root,节点,targetSum
来源： https://www.cnblogs.com/rsapaper/p/16184035.html