[LeetCode][JavaScript]Lowest Common Ancestor of a Binary Search Tree

Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /                  ___2__          ___8__
   /      \        /         0      _4       7       9
         /           3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

2轮遍历，记录下p和q的路径，比较路径输出不同的那个节点。

 1 /**
 2  * Definition for a binary tree node.
 3  * function TreeNode(val) {
 4  *     this.val = val;
 5  *     this.left = this.right = null;
 6  * }
 7  */
 8 /**
 9  * @param {TreeNode} root
10  * @param {TreeNode} p
11  * @param {TreeNode} q
12  * @return {TreeNode}
13  */
14 var lowestCommonAncestor = function(root, p, q) {
15     var pathP = [];
16     var pathQ = [];
17     findLCA(root, p.val, pathP);
18     findLCA(root, q.val, pathQ);
19     var len = Math.min(pathP.length, pathQ.length);
20     for(var i = 0; i <= len; i++){
21         if(!pathP[i] || !pathQ[i] || pathP[i].val !== pathQ[i].val){
22             return pathP[--i];
23         }
24     }
25
26     function findLCA(node, target, list){
27         list.push(node);
28         if(target < node.val){
29             findLCA(node.left, target, list);
30         }else if(target > node.val){
31             findLCA(node.right, target, list);
32         }
33     }
34 };

第一遍写了这么个东西。

先看左子树里有没有其中一个，如果有先记下，然后看右子树里有没有，加上左子树的结果判断是否找到，

最后加上目前节点的val，看是否找到。

虽然一轮遍历就可以知道结果，但是超级复杂，写了好长，没有用到BST的性质...

 1 /**
 2  * Definition for a binary tree node.
 3  * function TreeNode(val) {
 4  *     this.val = val;
 5  *     this.left = this.right = null;
 6  * }
 7  */
 8 /**
 9  * @param {TreeNode} root
10  * @param {TreeNode} p
11  * @param {TreeNode} q
12  * @return {TreeNode}
13  */
14 var lowestCommonAncestor = function(root, p, q) {
15     var target = [], res, isFound = false;
16     target.push(p.val);
17     target.push(q.val);
18     findLCA(root);
19     return res;
20
21     function findLCA(node){
22         if(isFound){
23             return [];
24         }
25
26         var candidates = [];
27         if(node.left !== null){
28             var left = findLCA(node.left);
29             left = compareVal(left);
30             if(left === true){
31                 res = node.left;
32                 isFound = true;
33                 return [];
34             }else if(left.length === 1){
35                 candidates= left;
36             }
37         }
38
39         if(node.right !== null && !isFound){
40             var right = findLCA(node.right);
41             if(right && right.length === 1){
42                 candidates.push(right[0]);
43             }
44             right = compareVal(candidates);
45             if(right === true){
46                 res = node;
47                 isFound = true;
48                 return [];
49             }else if(right.length === 1){
50                 candidates = right;
51             }
52         }
53
54         if(!isFound){
55             candidates.push(node.val);
56             var curr = compareVal(candidates);
57             if(curr === true){
58                 res = node;
59                 isFound = true;
60                 return [];
61             }
62             return curr;
63         }
64
65     }
66     function compareVal(a, b){
67         if(typeof a === "object"){
68             if(a.length === 2){
69                 b = a[1];
70                 a = a[0];
71             }else if(a.length === 1){
72                 a = a[0];
73             }
74         }
75         var index1 = target.indexOf(a);
76         var index2 = target.indexOf(b);
77         if(index1 !== -1 && index2 !== -1 && index1 !== index2){
78             return true;
79         }else{
80             if(index1 !== -1){
81                 return [a];
82             }else if(index2 !== -1){
83                 return [b];
84             }
85         }
86         return [];
87     }
88 };