LintCode-88.最近公共祖先

最近公共祖先

给定一棵二叉树，找到两个节点的最近公共父节点(LCA)。

最近公共祖先是两个节点的公共的祖先节点且具有最大深度。

注意事项

假设给出的两个节点都在树中存在

样例

对于下面这棵二叉树



LCA(3, 5) = 4

LCA(5, 6) = 7

LCA(6, 7) = 7

标签

LintCode 版权所有 领英 二叉树 脸书

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 search tree.
     * @param A and B: two nodes in a Binary.
     * @return: Return the least common ancestor(LCA) of the two nodes.
     */
    TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *A, TreeNode *B) {
        // write your code here
        TreeNode * ancestor = NULL;
        if(root==NULL || A==NULL || B==NULL)
            return ancestor;

        vector<TreeNode *> pathA, pathB;
        searchTree(root, A, pathA);
        searchTree(root, B, pathB);

        int sizeA=pathA.size(), sizeB=pathB.size();
        int size = sizeA>sizeB ? sizeB : sizeA, i=0;
        bool find = false;

        for(i=0; i<size; i++) {
            if(pathA[i] != pathB[i]) {
                find = true;
                return ancestor;
            }
            else
                ancestor = pathA[i];
        }
        if(find == false)
            ancestor = pathA[size-1];
        return ancestor;
    }

    bool searchTree(TreeNode *root, TreeNode *node, vector<TreeNode *> &path) {
        if(root==NULL || node==NULL) {
            return false;
        }

        path.push_back(root);

        //  找到了
        if(root->val == node->val) {
            return true;
        }

        if(root->left != NULL) {
            if(searchTree(root->left, node, path)) {
                return true;
            }
        }
        if(root->right != NULL) {
            if(searchTree(root->right, node, path)) {
                return true;
            }
        }

        //回溯
        path.pop_back();
        return false;
    }
};