113. 路径总和 II

题目描述

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

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,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：[]

示例 3：

输入：root = [1,2], targetSum = 0
输出：[]

提示：

树中节点总数在范围 [0, 5000] 内
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000

解法一：DFS

如下解法会导致每一条满足条件的路径被添加两次

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, targetSum int) [][]int {
    var path []int
    var ans [][]int
    var dfs func(root *TreeNode, curSum int)
    dfs = func(root *TreeNode, curSum int) {
        if nil == root {
            if curSum == targetSum {
                ans = append(ans, append([]int(nil), path...))
            }
            return
        }
        path = append(path, root.Val)
        dfs(root.Left, curSum + root.Val)
        dfs(root.Right, curSum+root.Val)
        path = path[:len(path)-1]
    }
    dfs(root, 0)
    return ans
}

下面是正确代码：

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, targetSum int) [][]int {
    var ans [][]int
    var path []int
    var dfs func(root *TreeNode, curSum int)
    dfs = func(root *TreeNode, curSum int) {
        if nil == root {
            return
        }
        curSum += root.Val
        path = append(path, root.Val)
        defer func() {path = path[:len(path)-1]}()
        if root.Left == nil && root.Right == nil {
            if curSum == targetSum {
                ans = append(ans, append([]int(nil), path...))
            }
            return
        }
        dfs(root.Left, curSum)
        dfs(root.Right, curSum)
    }
    dfs(root, 0)
    return ans
}

解法二：BFS

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, targetSum int) [][]int {
    var ans [][]int
    if nil == root {
        return ans
    }
    parent := make(map[*TreeNode]*TreeNode)
    var buildPath func(node *TreeNode) []int
    buildPath = func(node *TreeNode) []int {
        var path []int
        for node != nil {
            path = append(path, node.Val)
            node = parent[node]
        }
        for i := 0; i < len(path)/2; i++ {
            path[i], path[len(path)-1-i] = path[len(path)-1-i], path[i]
        }
        return path
    }
    parent[root] = nil
    type elem struct {
        node *TreeNode
        sum int
    }
    var queue []elem
    queue = append(queue, elem{node: root, sum: root.Val})
    for len(queue) > 0 {
        cur := queue[0]
        queue = queue[1:]
        if cur.node.Left == nil && cur.node.Right == nil {
            if cur.sum == targetSum {
                ans = append(ans, buildPath(cur.node))
            }
        }
        if cur.node.Left != nil {
            queue = append(queue, elem{node: cur.node.Left, sum: cur.sum+cur.node.Left.Val})
            parent[cur.node.Left] = cur.node
        }
        if cur.node.Right != nil {
            queue = append(queue, elem{node: cur.node.Right, sum: cur.sum+cur.node.Right.Val})
            parent[cur.node.Right] = cur.node
        }
    }
    return ans
}