To prove safety OFFER ---- 68-1, recent common ancestor binary search tree (js achieve)

Topic
Given a binary search tree, find the nearest common ancestor of the two specified nodes of the tree.

Baidu Encyclopedia recent common ancestor as defined: "For two nodes of the root of the tree T p, q, is represented as a common ancestor node x, that x is p, q ancestor depth as large as possible and x (a node can be its own ancestors). "

For example, given the following binary search tree: root = [6,2,8,0,4,7,9, null, null, 3,5]

Input: root = [6,2,8,0,4,7,9, null, null, 3,5], p = 2, q = 8
Output: 6
Explanation: nodes 2 and 8 is the common ancestor 6.

Input: root = [6,2,8,0,4,7,9, null, null, 3,5], p = 2, q = 4
Output: 2
Explanation: nodes 2 and 4 are common ancestor 2, because by definition a recent common ancestor node can be a node itself.

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Thinking

The characteristics of the binary search tree, p, q any one of equal to the parent node, returns, p, q are smaller than the parent node values ​​are in the left tree continues, is greater than the parent value is continued in the right tree, a parent node is greater than, a value smaller than the parent node, the parent node returns

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/**
 * 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}
 */

// 递归版本
var lowestCommonAncestor = function (root, p, q) {
        if (!root) {
          return null
        }
        if (p.val === root.val) {
          return p
        } else if (q.val === root.val) {
          return q
        } else if (root.val > p.val && root.val > q.val) {
          return lowestCommonAncestor(root.left, p, q)
        } else if (root.val < p.val && root.val < q.val) {
          return lowestCommonAncestor(root.right, p, q)
        } else if ((root.val < p.val && root.val > q.val) || (root.val > p.val && root.val < q.val)) {
          return root
        }
};

// 迭代版本
var lowestCommonAncestor = function (root, p, q) {
    if (!root) {
      return null
    }
    while (root) {
      if (p.val === root.val) {
        return p
      } else if (q.val === root.val) {
        return q
      } else if (p.val < root.val && q.val < root.val) {
        root = root.left
      } else if (p.val > root.val && q.val > root.val) {
        root = root.right
      } else if ((p.val > root.val && q.val < root.val) || (p.val < root.val && q.val > root.val)) {
        return root
      }
    }
    return null
};