Lowest Common Ancestor of a Binary Tree

题目连接

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

Common Ancestor of a Binary Tree

Description

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

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).”

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

先找到节点到根节点的路径(两个节点两条链)，再倒着从这两条链找，相遇的第一个节点即为所求。。 
当然，求LCA还有更高效的，倍增，tarjan离线，。。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
	typedef TreeNode* T;
	typedef vector<T> vec;
public:
	T lowestCommonAncestor(T root, T p, T q) {
		if (!root || !p || !q) return NULL;
		A.clear(), B.clear(), C.clear();
		dfs(root, p, q, A, B, C);
		if (B.empty() || C.empty()) return NULL;
		reverse(B.begin(), B.end());
		reverse(C.begin(), C.end());
		for (auto &r : B) {
			vec::iterator it = find(C.begin(), C.end(), r);
			if (it != C.end()) return r;
		}
		return NULL;
	}
private:
	void dfs(T x, T p, T q, vec &A, vec &B, vec &C) {
		if (!x) return;
		A.push_back(x);
		if (x == p) B = A;
		if (x == q) C = A;
		dfs(x->left, p, q, A, B, C);
		dfs(x->right, p, q, A, B, C);
		A.pop_back();
	}
	vec A, B, C;
};