Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

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).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5 

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

来源： <https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/>

  

思路，先判断入口是否有非法输入。

 

1 如果某一个root==p || root == q，那么LCA肯定是root（因为是top down，LCA肯定在root所囊括的树上，而root又是p q其中一个节点了，那么另外一个节点肯定在root之下，那么root就是LCA），那么返回root

 

2 如果root<min(p, q)，那么LCA肯定在右子树上，那么递归

 

3 如果max(p, q)<root，那么LCA肯定在左子树上，那么递归

 

4 如果p<root<q，那么root肯定为LCA

/**
 * 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==p||root==q)
            return root;
        if(p->val>q->val)
            swap(p,q);
        if(root->val > p->val && root->val < q->val)
            return root;
        TreeNode *node;
        if(root->val > max(p->val,q->val))
            node=lowestCommonAncestor(root->left,p,q);
        if(root->val < min(p->val,q->val))
            node=lowestCommonAncestor(root->right,p,q);
        
        return node;
    }
   
};

来自为知笔记(Wiz)