LeetCode-112. Path Sum

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问题描述

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.

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：递归
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/*
 * @Description: 112. Path Sum
 * @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: 递归解法
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {boolean}
 */
const hasPathSum = function (root, targetSum) {
  if (!root) {
    return false
  }
  if (root.left === null && root.right === null && targetSum === root.val) {
    return true
  }
  return (hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val))
}
方法二：DFS
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/*
 * @Description: 112. Path Sum
 * @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
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {boolean}
 */
const hasPathSum = function (root, targetSum) {
  if (!root) {
    return false
  }

  let nodeStack = [root]
  let valStack = [root.val]

  while (nodeStack.length) {
    let cur = nodeStack.pop()
    let curVal = valStack.pop()

    if (cur.left === null && cur.right === null && curVal === targetSum) {
      return true
    }
    if (cur.left) {
      nodeStack.push(cur.left)
      valStack.push(curVal + cur.left.val)
    }
    if (cur.right) {
      nodeStack.push(cur.right)
      valStack.push(curVal + cur.right.val)
    }
  }
  return false
}