leetCode(39):Lowest Common Ancestor of a Binary Tree 分类： leetCode 2015-07-17 10:03 114人阅读 评论(0) 收藏

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.

首先找出自根结点到两个结点的路径，并保存，然后找这两条路径最后一个相同的结点

/**
 * 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:
    bool getPath(TreeNode* root, TreeNode* node, list<TreeNode*>& path)
    {
    	if (root == node)
    	{
    		path.push_back(root);
    		return true;
    	}
    	path.push_back(root);
    
    	TreeNode* tmp = root;
    
    	bool found_left = false;
    	bool found_right = false;
    
    	if (tmp->left)
    		found_left = getPath(tmp->left, node, path);
    	if (!found_left && tmp->right)
    	{//右子树中没有则在其左子树中寻找
    		found_right = getPath(tmp->right, node, path);
    	}
    	
    	if (!found_left && !found_right)//左右子树中都未找到，则路径错误
    		path.pop_back();
    
    	return found_left || found_right;//返回查找结果
    }

    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root == NULL)
    		return NULL;
    	list<TreeNode*> path1, path2;
    	getPath(root, p, path1);
    	getPath(root, q, path2);
    	list<TreeNode*>::iterator iter1, iter2;
    	iter1 = path1.begin();
    	iter2 = path2.begin();
    	TreeNode* plast=NULL;
    	while (iter1 != path1.end() && iter2 != path2.end())
    	{
    		if (*iter1 == *iter2)
    		{//查找最后一个相同的结点，在此之前，都是相同的
    			plast = *iter1;
    		}
    		iter1++;
    		iter2++;
    	}
    	return plast;
    }
};