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

According to the : “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;};