[LeetCode]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).”

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

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.

［思路］

深度优先搜索搜寻节点，存储节点路径。在两个路径中找到公共节点。注意不是二叉搜索树。

［代码］

/**
 * 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 {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        vector<TreeNode*> path1;
        vector<TreeNode*> path2;
        TreeNode* ret;
        FindPath(root,path1,p);
        FindPath(root,path2,q);
        for(int i=0 ;i<min(path1.size(),path2.size());++i){
            if(path1[i]==path2[i])
                ret = path1[i];
        }
        return ret;
    }
    void FindPath(TreeNode *root,vector<TreeNode*> &path,TreeNode *find){
        if(root == nullptr){
            return;
        }
        if(root == find){
            path.push_back(find);
            return;
        }
        path.push_back(root);
        FindPath(root->left,path,find);
        FindPath(root->right,path,find);
        path.pop_back();
    }
};