leetcode - Lowest Common Ancestor of a Binary Tree

题目：

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.

分析：

求二叉树节点最低公共祖先的题目，应该做到以下两点：

1、时间复杂度O(n)，只需遍历一遍二叉树。

2、要判断输入的两个节点是否在二叉树中。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */  struct node {     TreeNode* res;     bool bp,bq;     node(TreeNode* n,bool x,bool y):res(n),bp(x),bq(y){} }; class Solution {public:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)     {         if(!root || !p || !q)            return nullptr;        return lowestCommonAncestor_core(root,p,q).res;    }     node lowestCommonAncestor_core(TreeNode* root, TreeNode* p, TreeNode* q)      {         if(root==nullptr)            return node(nullptr,false,false);         if(root==p)         {                          return hasnode(root,q)?node(root,true,true):node(root,true,false);         }         if(root==q)         {                          return hasnode(root,p)?node(root,true,true):node(root,false,true);         }         if(root->left)         {             auto r=lowestCommonAncestor_core(root->left,p,q);             if(r.bp && r.bq)                return r;            else if(!r.bp && r.bq)            {                if(hasnode(root->right,p))                    return node(root,true,true);                else                    return node(nullptr,false,true);            }            else if(r.bp && !r.bq)            {                if(hasnode(root->right,q))                    return node(root,true,true);                else                    return node(nullptr,true,false);            }         }                 return lowestCommonAncestor_core(root->right,p,q);                  }    bool hasnode(TreeNode* root, TreeNode* p)    {        if(root==p)            return true;        if(root==nullptr)            return false;        return hasnode(root->left,p) || hasnode(root->right,p);    }};