【LeetCode】236. Lowest Common Ancestor of a Binary Tree

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.

好巧，我在Lowest Common Ancestor of a Binary Search Tree的解法一，

就是这题的解法。

深度遍历到节点p时，栈中的所有节点即为p的从根开始的祖先序列。

因此只需要比较p、q祖先序列中最后一个相同的祖先即可。

/**
 * 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) {
        // special cases
        if(root == NULL)
            return NULL;
        if(p == root || q == root)
            return root;
        if(p == q)
            return p;
            
        vector<TreeNode*> vp;
        vector<TreeNode*> vq;
        stack<TreeNode*> stk;
        unordered_map<TreeNode*, bool> m;   //visited
        stk.push(root);
        m[root] = true;
        while(!stk.empty())
        {
            TreeNode* top = stk.top();
            if(top->left && m[top->left] == false)
            {
                stk.push(top->left);
                m[top->left] = true;
                if(top->left == p)
                {
                    vp = stkTovec(stk);
                    if(!vq.empty())
                        break;
                }
                if(top->left == q)
                {
                    vq = stkTovec(stk);
                    if(!vp.empty())
                        break;
                }
                continue;
            }
            if(top->right && m[top->right] == false)
            {
                stk.push(top->right);
                m[top->right] = true;
                if(top->right == p)
                {
                    vp = stkTovec(stk);
                    if(!vq.empty())
                        break;
                }
                if(top->right == q)
                {
                    vq = stkTovec(stk);
                    if(!vp.empty())
                        break;
                }
                continue;
            }
            stk.pop();
        }
        int i = 0;
        for(; i < vp.size() && i < vq.size(); i ++)
        {
            if(vp[i] != vq[i])
                break;
        }
        return vp[i-1];
    }
    vector<TreeNode*> stkTovec(stack<TreeNode*> stk)
    {
        vector<TreeNode*> v;
        while(!stk.empty())
        {
            TreeNode* top = stk.top();
            stk.pop();
            v.push_back(top);
        }
        reverse(v.begin(), v.end());
        return v;
    }
};