[LeetCode]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.

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简单题，不过加上思考和编码的时间也大概有半小时了吧。。。

思路就是使用dfs，分别dfs 寻找两个数，记录下从root 到目标node 经过了哪些node，如果在某一次dfs 中发现了存在的node 就说明那个node就是LCA。关键在于，要在递归后检查，而不是在递归前，这样才能保证是lowest 的，因为整个检查过程是倒桩的。如果是在递归前检查，就变成最高祖先了(root)。

/**
 * 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) {
    var _lca = null;
    var _set = {};
    function dfs(node, a, set) {
        if (!node) return;
        if (node.val == a) {
            if (set[node.val] != void 0 && _lca === null) _lca = node;
            else set[node.val] = 1;
            return a;
        }
        var l_find = dfs(node.left, a, set);
        if (l_find == a) {
            if (set[node.val] != void 0 && _lca === null) _lca = node;
            else set[node.val] = 1;
            return a;
        }
        var r_find = dfs(node.right, a, set);
        if (r_find == a) {
            if (set[node.val] != void 0 && _lca === null) _lca = node;
            else set[node.val] = 1;
            return a;
        }
    }

    dfs(root, p.val, _set);
    dfs(root, q.val, _set);

    return _lca;
};