[LeetCode]236.Lowest Common Ancestor of a Binary Tree

题目

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    4Another example is LCA of nodes 5 and 4 is 5,  since a node can be a descendant of itself according to the LCA definition.

代码

/*---------------------------------------*   日期：2015-07-13*   作者：SJF0115*   题目: 236.Lowest Common Ancestor of a Binary Tree*   网址：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/*   结果：AC*   来源：LeetCode*   博客：-----------------------------------------*/#include <iostream>#include <vector>using namespace std;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 || p == nullptr || q == nullptr){            return nullptr;        }//if        vector<TreeNode*> path1;        bool isFind = Path(root,p,path1);        // 没有P节点        if(!isFind){            return nullptr;        }//if        vector<TreeNode*> path2;        isFind = Path(root,q,path2);        if(!isFind){            return nullptr;        }//if        int size1 = path1.size();        int size2 = path2.size();        // 求最近祖先        TreeNode* node = nullptr;        for(int i = 0,j = 0;i <= size1 && j <= size2;++i,++j){            if((i == size1 || j == size2) || path1[i] != path2[j]){                node = path1[i-1];                break;            }//if        }//for        return node;    }private:    // 从根节点到node节点的路径    bool Path (TreeNode* root,TreeNode* node,vector<TreeNode*> &path) {        path.push_back(root);        if(root == node) {            return true;        }//if        bool isExits = false;        // 左子树        if(root->left) {            isExits = Path(root->left,node,path);        }//if        // 右子树        if(!isExits && root->right) {            isExits = Path(root->right,node,path);        }//if        if(!isExits) {            path.pop_back();        }//if        return isExits;    }};int main(){    Solution s;    TreeNode* root = new TreeNode(3);    TreeNode* node1 = new TreeNode(0);    TreeNode* node2 = new TreeNode(1);    TreeNode* node3 = new TreeNode(2);    TreeNode* node4 = new TreeNode(4);    TreeNode* node5 = new TreeNode(5);    TreeNode* node6 = new TreeNode(6);    TreeNode* node7 = new TreeNode(7);    TreeNode* node8 = new TreeNode(8);    root->left = node5;    root->right = node2;    node5->left = node6;    node5->right = node3;    node3->left = node7;    node3->right = node4;    node2->left = node1;    node2->right = node8;    TreeNode* node = s.lowestCommonAncestor(root,node7,node1);    if(node != nullptr){        cout<<node->val<<endl;    }//if    else{        cout<<"nullptr"<<endl;    }//else    return 0;}

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