Lowest Common Ancestor III

Lowest Common Ancestor III -1.png

Lowest Common Ancestor III -2.png

解題思路 ：

跟普通找 LCA 的差別就是 題目給的 ＡＢ兩點可能根本不在樹裡面 所以另外設定 boolean 檢查是否真的找到 A 或 B 點 如果有一點沒找到的話 就從找到的另一點往下搜尋看看是否有包含沒找到的點

C++ code :

 /**
 
 
 
  
Definition of TreeNode:
 
  
class TreeNode {
 
  
public:
 
  
int val;
 
  
TreeNode *left, *right;
 
  
TreeNode(int val) {
 
  
    this->val = val;
 
  
    this->left = this->right = NULL;
 
  
}
 
  
}
 */
 
 
 
 
class Solution {
 
 
public:
 
 
/**
 * @param root: The root of the binary tree.
 * @param A and B: two nodes
 * @return: Return the LCA of the two nodes.
 */
TreeNode *LCA(TreeNode* root, TreeNode* A, TreeNode* B, bool &findA, bool &findB)
{
    if(!root) return root;
    if(root == A){
        findA = true;
        return root;
    }
    if(root == B){
        findB = true;
        return root;
    }
    TreeNode *left = LCA(root->left, A, B, findA, findB);
    TreeNode *right = LCA(root->right, A, B, findA, findB);
    if(left && right) return root;
    return left? left : right;
}

bool search(TreeNode *res, TreeNode *n)
{
    if(res == nullptr) return false;
    if(res == n) return true;
    return search(res->left, n) || search(res->right, n);
}

TreeNode *lowestCommonAncestor3(TreeNode* root, TreeNode* A, TreeNode* B) {
    // write your code here
    bool findA = false, findB = false;
    TreeNode *res = LCA(root, A, B, findA, findB);
    bool check = true;
    if(!findA || !findB)
    {
        if(res == A) check = search(res, B);
        else check = search(res, A);
    }
    return check? res : nullptr;
}
 
 
};