LeetCode-113. Path Sum II

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问题描述

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.

A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.

Example :

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Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22

Constraints:

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

解答

方法一：DFS迭代写法
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/*
 * @Description: 113. Path Sum II
 * @Author: libk
 * @Github: https://github.com/libk
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @description: DFS迭代写法
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {number[][]}
 */
const pathSum = function(root, targetSum) {
  if (!root) {
    return []
  }
  let res = []
  let nodeStack = [root]
  let valStack = [root.val]
  let pathStack = [[root.val]]

  while (nodeStack.length) {
    let cur = nodeStack.pop()
    let curVal = valStack.pop()
    // 当前节点经过的路径
    let curPath = pathStack.pop()

    if (cur.left === null && cur.right === null && curVal === targetSum) {
      res.push(curPath)
    }
    if (cur.left) {
      let leftPath = curPath.slice(0)
      leftPath.push(cur.left.val)
      pathStack.push(leftPath)
      nodeStack.push(cur.left)
      valStack.push(curVal + cur.left.val)
    }
    if (cur.right) {
      let rightPath = curPath.slice(0)
      rightPath.push(cur.right.val)
      pathStack.push(rightPath)
      nodeStack.push(cur.right)
      valStack.push(curVal + cur.right.val)
    }
  }
  return res
}
方法二：DFS递归写法
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/*
 * @Description: 113. Path Sum II
 * @Author: libk
 * @Github: https://github.com/libk
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @description: DFS递归写法
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {number[][]}
 */
const pathSum = function(root, targetSum) {
  let res = []
  let path = []

  const dfs = (cur, curSum) => {
    if (!cur) {
      return
    }
    path.push(cur.val)
    curSum -= cur.val
    if (cur.left === null && cur.right === null && curSum === 0) {
      res.push(path.slice(0))
    }
    dfs(cur.left, curSum)
    dfs(cur.right, curSum)
    path.pop()
  }

  dfs(root, targetSum)
  return res
}