leetcode 236. Lowest Common Ancestor of a Binary Tree 最近公告祖先LCA + 二叉树 + 深度优先遍历DFS

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.

和上一道题一样，直接DFS深度优先遍历即可。

建议和这道题leetcode 235. Lowest Common Ancestor of a Binary Search Tree 最近公共祖先 + BST 一起学习。

代码如下：

/*class TreeNode
{
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}*/

public class Solution
{
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q)
{
//发现目标节点则通过返回值标记该子树发现了某个目标结点
if(root == null || root == p || root == q)
return root;

//查看左子树中是否有目标结点，没有为null
TreeNode left = lowestCommonAncestor(root.left, p, q);
//查看右子树是否有目标节点，没有为null
TreeNode right = lowestCommonAncestor(root.right, p, q);

//都不为空，说明做右子树都有目标结点，则公共祖先就是本身
if(left!=null&&right!=null)
return root;

//如果发现了目标节点，则继续向上标记为该目标节点
return left == null ? right : left;
}
}

下面是C++的做法，这里还是DFS深度优先搜索来遍历二叉树来找到LCA的，值得学习

代码如下：

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <string>
#include <map>

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