【一天一大 lee】二叉搜索树的最近公共祖先 (难度:简单) - Day2020092

输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
输出: 6
解释: 节点 2 和节点 8 的最近公共祖先是 6。

输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
输出: 2
解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */

/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function(root, p, q) {
  let pathP = getPath(root, p)
  let pathQ = getPath(root, q)
  let _result = null
  for (let i = 0; i < pathP.length && i < pathQ.length; ++i) {
    // 找到最后一个相同的节点
    if (pathP[i] == pathQ[i]) {
      _result = pathP[i]
      continue
    } else {
      break
    }
  }
  // 生成从根节点遍历的到目标节点的路径数组
  function getPath(root, target) {
    let path = []
    let node = root
    while (node != target) {
      path.push(node)
      if (target.val < node.val) {
        node = node.left
      } else {
        node = node.right
      }
    }
    path.push(node)
    return path
  }
  return _result
}

     ...     ...     ...
     /         \     /  \
    p           p      p    q
   /             \
 ...             ...
 /                 \

var lowestCommonAncestor = function(root, p, q) {
  let node = root
  while (node) {
    if (p.val < node.val && q.val < node.val) {
      node = node.left
    } else if (p.val > node.val && q.val > node.val) {
      node = node.right
    } else {
      break
    }
  }
  return node
}