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https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/
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;
};
Lowest Common Ancestor of a Binary Tree
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原文地址：http://www.cnblogs.com/GadyPu/p/5020671.html
