码迷,mamicode.com
首页 > 其他好文 > 详细

[LeetCode]Lowest Common Ancestor of a Binary Tree

时间:2015-07-26 11:11:32      阅读:136      评论:0      收藏:0      [点我收藏+]

标签:

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]

深度优先搜索搜寻节点,存储节点路径。在两个路径中找到公共节点。注意不是二叉搜索树。

[代码]

/**
 * 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<TreeNode*> path1;
        vector<TreeNode*> path2;
        TreeNode* ret;
        FindPath(root,path1,p);
        FindPath(root,path2,q);
        for(int i=0 ;i<min(path1.size(),path2.size());++i){
            if(path1[i]==path2[i])
                ret = path1[i];
        }
        return ret;
    }
    void FindPath(TreeNode *root,vector<TreeNode*> &path,TreeNode *find){
        if(root == nullptr){
            return;
        }
        if(root == find){
            path.push_back(find);
            return;
        }
        path.push_back(root);
        FindPath(root->left,path,find);
        FindPath(root->right,path,find);
        path.pop_back();
    }
};


版权声明:本文为博主原创文章,未经博主允许不得转载。

[LeetCode]Lowest Common Ancestor of a Binary Tree

标签:

原文地址:http://blog.csdn.net/ciaoliang/article/details/47065771

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!