LeetCode236. Lowest Common Ancestor of a Binary Tree（C++）

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]

 

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

• All of the nodes' values will be unique.
• p and q are different and both values will exist in the binary tree.

方法一：沿用236的方法，如果该节点为p或q，说明该节点为所求的LCA，否则在其左子树和右子树寻找他们的LCA，如果左右子树都找到LCA，只有可能该节点就是LCA，否则，一定有一个子树求到的LCA为空，而不为空的那个子树的LCA即为我们所求结果。

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == NULL)
            return root;
        if(root == p || root == q)
            return root;
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        if(left != NULL && right != NULL)
            return root;
        else if(left != NULL)
            return left;
        else if(right != NULL)
            return right;
        return NULL;
    }
};

解题思路：先遍历找到两个节点从根节点到该节点的路径，用0、1表示走左子树还是右子树。然后再寻找两个路径第一个不一样的分叉。

class Solution {
public:
    int flag = 0;
    bool findflag = false;
    vector<int>temppath[2],anspath[2];
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        findvalue(root,p);
        flag = 1;
        findflag = false;
        findvalue(root,q);
        TreeNode *ans = root;
        int cnt = 0, len = min(anspath[0].size(), anspath[1].size());
        while(cnt < len && anspath[0][cnt] == anspath[1][cnt]){
            if(anspath[0][cnt] == 0)
                ans = ans->left;
            else
                ans = ans->right;
            cnt++;
        }
        return ans;
    }
    void findvalue(TreeNode *root, TreeNode *x){
        if(root == NULL || findflag == true)
            return;
        if(root->val == x->val){
            anspath[flag] = temppath[flag];
            findflag = true;
            return;
        }
        temppath[flag].push_back(0);
        findvalue(root->left, x);
        temppath[flag].pop_back();
        
        temppath[flag].push_back(1);
        findvalue(root->right, x);
        temppath[flag].pop_back();
    }
};