LintCode 近期公共祖先

中等 近期公共祖先 查看执行结果 

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通过

给定一棵二叉树，找到两个节点的近期公共父节点(LCA)。

近期公共祖先是两个节点的公共的祖先节点且具有最大深度。

您在真实的面试中是否遇到过这个题？ 

Yes

例子

对于以下这棵二叉树

  4
 / 
3   7
   / 
  5   6

LCA(3, 5) = 4

LCA(5, 6) = 7

LCA(6, 7) = 7







/**
 * 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 search tree.
     * @param A and B: two nodes in a Binary.
     * @return: Return the least common ancestor(LCA) of the two nodes.
     */
    bool   postOrder(TreeNode *root, TreeNode *A) {
    if (root == nullptr)
        return false;
    if (root == A) {
        return true;
    }
    return (postOrder(root->left,A)|| postOrder(root->right,A));
}
    TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *A, TreeNode *B) {
        // write your code here
        if (root == nullptr )
        return nullptr;
    bool ra = postOrder(root->right,A);
    bool rb = postOrder(root->right,B);
    bool la = postOrder(root->left,A);
    bool lb = postOrder(root->left,B);
        if (ra && rb) {
            return lowestCommonAncestor (root->right,A,B);
        } else if (la && lb) {
            return lowestCommonAncestor(root->left,A,B);
        } else {
            return root;
        }
    }

};