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leetcode_236——Lowest Common Ancestor of a Binary Tree(tree,后序遍历)

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Lowest Common Ancestor of a Binary Tree

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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.

 

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      这道题采用二叉树的后序遍历来做,当左右子树返回0时,判断根结点,若为不等则返回0,若等返回1;
                                                  当左右子树一个返回1,一个返回0,判断根结点,若等说明已经找到了,其实已经不用递归了,赋值,然后返回0(不让后再找到),若不等,则返回1.
                                                   当左右子树返回0时,则判断根结点,若等则返回1,若不等则返回0.
#include<iostream>
using namespace std;

struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;
     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 };

TreeNode* L=NULL;

int digui(TreeNode* root,TreeNode* p,TreeNode* q)
{
	if(root->left==NULL&&root->right==NULL)
	{
		if(root==p||root==q)
			return 1;
		else
			return 0;
	}
	int left=0,right=0;
	if(root->left!=NULL)
		left=digui(root->left,p,q);
	if(root->right!=NULL)
		right=digui(root->right,p,q);
	if(left+right==0)
	{
		if(root==p||root==q)
			return 1;
		else
			return 0;
	}
	else if(left+right==1)
	{
		if(root==p||root==q)
		{
			L=root;
			return 0;
		}
		else
			return 1;
	}
	else
	{
		L=root;
		return 0;
	}
}

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
	digui(root,p,q);
	return L;
}
int main()
{

}

  

leetcode_236——Lowest Common Ancestor of a Binary Tree(tree,后序遍历)

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原文地址:http://www.cnblogs.com/yanliang12138/p/4658140.html

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