1008. 前序遍历构造二叉搜索树

题目

返回与给定前序遍历 preorder 相匹配的二叉搜索树（binary search tree）的根结点。

二叉搜索树每个节点都满足以下规则

• node.left 的任何后代，值总 < node.val
• node.right 的任何后代，值总 > node.val

题解

递归

/**
 * 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)
 * }
 */
/**
 * @param {number[]} preorder
 * @return {TreeNode}
 */
var bstFromPreorder = function(preorder) {
  if (!preorder.length) return null;
  let root = new TreeNode(preorder[0]);
  let index = 1;
  while (index < preorder.length && preorder[index] < preorder[0]) {
    index++;
  }
  root.left = bstFromPreorder(preorder.slice(1, index));
  root.right = bstFromPreorder(preorder.slice(index));
  return root;
};

迭代

/**
 * 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)
 * }
 */
/**
 * @param {number[]} preorder
 * @return {TreeNode}
 */
var bstFromPreorder = function(preorder) {
  if (!preorder.length) return null;
  let root = new TreeNode(preorder[0]);
  let s = [root];
  for (let i = 1; i < preorder.length; ++i) {
    let node = s[s.length - 1];
    let child = new TreeNode(preorder[i]);
    while (s.length && s[s.length - 1].val < child.val) {
      node = s.pop();
    }
    if (node.val < child.val) node.right = child;
    else node.left = child;
    s.push(child);
  }
  return root;
};