236. Lowest Common Ancestor of a Binary Tree-蒲公英云

236. 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 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]

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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.

题意：寻找二叉树上，两个结点的公共祖先。

方法一：递归

/**
 * 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)
    {        
       if (root == nullptr)
            return nullptr;
       if(p == root || q == root)
           return root;
       TreeNode* left = lowestCommonAncestor(root->left,p,q);
       TreeNode* right =lowestCommonAncestor(root->right,p,q);
       if(left == nullptr)
           return right;
       if(right == nullptr)
           return left;
        return root;        
    }    
};

方法二：计算结点p,q到根节点的路径，比较这两个路径，返回公共祖先结点。

/**
 * 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)
    {        
        TreeNode* t1 = root;
        TreeNode* t2 = root;
        vector<TreeNode*> path1,path2;
        createPath(t1, p, path1);
        createPath(t2, q, path2);
        int len = (path1.size() < path2.size()? path1.size():path2.size());
        int i = 0;
        for (; i < len; i++)
        {
            if(path1[i] != path2[i])
                break;
        }
        return  path1[i-1];
    }
    bool createPath(TreeNode* root, TreeNode* key, vector<TreeNode*>& path)
    {
        if (root == nullptr)
            return false;
        path.push_back(root);
        if (root == key)
            return true;     
        if (createPath(root->left, key, path) || createPath(root->right, key, path))
        {
            return true;
        }
        path.pop_back();
        return false;
    }
};