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;
};