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

最新推荐文章于 2022-11-16 10:43:06 发布

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本文介绍了一种解决二叉树中寻找两节点最近公共祖先(LCA)问题的方法。通过递归查找从根节点到目标节点的路径，并利用路径信息找到最近的共同祖先节点。

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 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.
题意：
给定一棵二叉树，查找两个节点的最近祖先节点。
思路：
首先找出从根节点到这两个节点的路径，然后从根节点往下走，到达两个节点最后一个相同的位置就是其最近的祖先节点。
代码如下：

/**
 * 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)
    {
        if(root == NULL || p == NULL || q == NULL) return NULL;
        vector<TreeNode*> pv, qv;
        if(!(getPath(root, p, pv) && getPath(root, q, qv)))return NULL;
        return getCommonAncestor(pv, qv);
    }
    bool getPath(TreeNode* root, TreeNode* p, vector<TreeNode*>& elements)
    {
        elements.push_back(root);
        if(root == p)
        {
            return true;
        }
        if(root->left != NULL)
        {
            if(getPath(root->left, p, elements))return true;
        }
        if(root->right != NULL)
        {
            if(getPath(root->right, p, elements))return true;
        }
        elements.pop_back();
        return false;
    }
    TreeNode* getCommonAncestor(vector<TreeNode*> pv, vector<TreeNode*> qv)
    {
        TreeNode* last = NULL;
        int index = 0;
        int sp = pv.size();
        int sq = qv.size();
        while(index < sp && index < sq && pv[index] == qv[index])
        {
            last = pv[index];
            index++;
        }
        return last;
    }
};