leecode_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 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;}