LeetCode-98. Validate Binary Search Tree

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问题描述

Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid BST is defined as follows:

  • The left subtree of a node contains only nodes with keys less than the node’s key.
  • The right subtree of a node contains only nodes with keys greater than the node’s key.
  • Both the left and right subtrees must also be binary search trees.

Example :

img

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Input: root = [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

解答

方法一：递归
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/*
 * @Description: 98. Validate Binary Search Tree
 * @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: 方法一：递归
 * @param {TreeNode} root
 * @return {boolean}
 */
const isValidBST = function (root) {
  const eventLoop = (cur, lower, upper) => {
    if (!cur) {
      return true
    }
    if (cur.val <= lower || cur.val >= upper) {
      return false
    }

    return eventLoop(cur.left, lower, cur.val) && eventLoop(cur.right, cur.val, upper)
  }

  return eventLoop(root, -Infinity, Infinity)
}
方法二：非递归中序遍历
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/**
 * @description: 方法二：非递归中序遍历
 * @param {TreeNode} root
 * @return {boolean}
 */
const isValidBST = function (root) {
  let pre = -Infinity
  let stack = []
  let cur = root

  while (cur || stack.length !== 0) {
    if (cur) {
      stack.push(cur)
      cur = cur.left
    } else {
      cur = stack.pop()
      if (cur.val <= pre) {
        return false
      }
      pre= cur.val
      cur = cur.right
    }
  }
  return true
}