leecode_236 Lowest Common Ancestor of a Binary Tree

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本文介绍了一种解决二叉树中寻找两节点最低公共祖先(LCA)问题的方法。通过广度优先搜索(BFS)和递归两种方式，找到了两节点路径上的最后一个共同节点。并提供了详细的C++实现代码。

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.

思路是找到两个点的路径，然后找最后一个相同的点：

c++实现：

/**
 * 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< vector<TreeNode*> > path=bfs(root,p,q);
        
        TreeNode* LCA;
        
        for (int i=0;i<min( path[1].size(), path[0].size() );i++){
            if (path[0][i]==path[1][i])
                LCA=path[0][i];
            else
               break;
        }
        
        return LCA;
    }
    
    vector< vector<TreeNode*> > bfs(TreeNode* root, TreeNode* p, TreeNode* q){
        queue<vector<TreeNode*>> que;
        vector<TreeNode*> path;
        vector< vector<TreeNode*> > res;
        path.push_back(root);
        if (root==p || root==q)
            res.push_back(path);
        que.push(path);
        
        while(!que.empty()){
            vector<TreeNode*> temp=que.front();
            que.pop();
            int len=temp.size();
            TreeNode* node=temp[len-1];
            
            
            
            
            if (node->left!=NULL){
                temp.push_back(node->left);
                que.push(temp);
                if (node->left==p || node->left==q)
                      res.push_back(temp);
                if (res.size()==2)
                      return res;
                temp.erase(temp.begin()+len);
            }
            
            if (node->right!=NULL){
                temp.push_back(node->right);
                que.push(temp);
                if (node->right==p || node->right==q)
                      res.push_back(temp);
                if (res.size()==2)
                      return res;
                temp.erase(temp.begin()+len);
            }
        }
        
        return res;
        
    }
};

递归法：

参考LeetCode Discuss

链接地址：https://leetcode.com/discuss/45386/4-lines-c-java-python-ruby

如果当前节点为空或者与目标节点之一相同，则返回当前节点

递归寻找p和q在左右子树中的位置

如果p和q分别位于root的左右两侧，则root为它们的LCA，否则为左子树或者右子树

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