Lowest Common Ancestor

Source

Given the root and two nodes in a Binary Tree. Find the lowest common ancestor(LCA) of the two nodes.

The lowest common ancestor is the node with largest depth which is the ancestor of both nodes.
Example
        4

    /     \

  3         7

          /     \

        5         6
For 3 and 5, the LCA is 4.

For 5 and 6, the LCA is 7.
For 6 and 7, the LCA is 7.

题解1 - 自底向上

初次接触这种题可能会没有什么思路，在没有思路的情况下我们就从简单例子开始分析！首先看看3和5，这两个节点分居根节点4的两侧，如果可以从子节点往父节点递推，那么他们将在根节点4处第一次重合；再来看看5和6，这两个都在根节点4的右侧，沿着父节点往上递推，他们将在节点7处第一次重合；最后来看看6和7，此时由于7是6的父节点，故7即为所求。从这三个基本例子我们可以总结出两种思路——自顶向下(从前往后递推)和自底向上(从后往前递推)。

顺着上述实例的分析，我们首先看看自底向上的思路，自底向上的实现用一句话来总结就是——如果遍历到的当前节点是 A/B 中的任意一个，那么我们就向父节点汇报此节点，否则递归到节点为空时返回空值。具体来说会有如下几种情况：

1、当前节点不是两个节点中的任意一个，此时应判断左右子树的返回结果。

• 若左右子树均返回非空节点，那么当前节点一定是所求的根节点，将当前节点逐层向前汇报。// 两个节点分居树的两侧
• 若左右子树仅有一个子树返回非空节点，则将此非空节点向父节点汇报。// 节点仅存在于树的一侧
• 若左右子树均返回NULL, 则向父节点返回NULL. // 节点不在这棵树中

2、当前节点即为两个节点中的一个，此时向父节点返回当前节点。
根据此递归模型容易看出应该使用先序/后序遍历来实现。

C++ Recursion From Bottom to Top

/**
 * 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) {
        // return either A or B or NULL
        if (NULL == root || root == A || root == B) return root;

        TreeNode *left = lowestCommonAncestor(root->left, A, B);
        TreeNode *right = lowestCommonAncestor(root->right, A, B);

        // A and B are on both sides
        if ((NULL != left) && (NULL != right)) return root;

        // either left or right or NULL
        return (NULL != left) ? left : right;
    }
};

Java

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @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.
     */
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
        if (root == null) return null;

        TreeNode lNode = lowestCommonAncestor(root.left, A, B);
        TreeNode rNode = lowestCommonAncestor(root.right, A, B);
        // root is the LCA of A and B
        if (lNode != null && rNode != null) return root;
        // root node is A/B(including the case below)
        if (root == A || root == B) return root;
        // return lNode/rNode if root is not LCA
        return (lNode != null) ? lNode : rNode;
    }
}

源码分析

结合例子和递归的整体思想去理解代码，在root == A || root == B后即层层上浮(自底向上)，直至找到最终的最小公共祖先节点。

最后一行return (NULL != left) ? left : right;将非空的左右子树节点和空值都包含在内了。

关于重复节点：由于这里比较的是元素地址，因此可以认为树中不存在重复元素，否则不符合树的数据结构。

题解 - 自底向上(计数器)

为了解决上述方法可能导致误判的情况，我们可以对返回结果添加计数器来解决。由于此计数器的值只能由子树向上递推，故应该用后序遍历。在类中添加私有变量较为方便。

定义pair<TreeNode *, int> result(node, counter)表示遍历到某节点时的返回结果，返回的node表示LCA 路径中的可能的最小节点，相应的计数器counter则表示目前和A或者B匹配的节点数，若计数器为2，则表示已匹配过两次，该节点即为所求，若只匹配过一次，还需进一步向上递推。表述地可能比较模糊，还是看看代码吧。

C++

/**
 * 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) {
        if ((NULL == A) || (NULL == B)) return NULL;

        pair<TreeNode *, int> result = helper(root, A, B);

        if (A != B) {
            return (2 == result.second) ? result.first : NULL;
        } else {
            return (1 == result.second) ? result.first : NULL;
        }
    }

private:
    pair<TreeNode *, int> helper(TreeNode *root, TreeNode *A, TreeNode *B) {
        TreeNode * node = NULL;
        if (NULL == root) return make_pair(node, 0);

        pair<TreeNode *, int> left = helper(root->left, A, B);
        pair<TreeNode *, int> right = helper(root->right, A, B);

        // return either A or B
        int count = max(left.second, right.second);
        if (A == root || B == root)  return make_pair(root, ++count);

        // A and B are on both sides
        if (NULL != left.first && NULL != right.first) return make_pair(root, 2);

        // return either left or right or NULL
        return (NULL != left.first) ? left : right;
    }
};

Java

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    private int count = 0;
    /**
     * @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.
     */
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
        TreeNode result = helper(root, A, B);
        if (A == B) {
            return result;
        } else {
            return (count == 2) ? result : null;
        }
    }

    private TreeNode helper(TreeNode root, TreeNode A, TreeNode B) {
        if (root == null) return null;

        TreeNode lNode = helper(root.left, A, B);
        TreeNode rNode = helper(root.right, A, B);
        // root is the LCA of A and B
        if (lNode != null && rNode != null) return root;
        // root node is A/B(including the case below)
        if (root == A || root == B) {
            count++;
            return root;
        }
        // return lNode/rNode if root is not LCA
        return (lNode != null) ? lNode : rNode;
    }
}

源码分析

在A == B时，计数器返回1的节点即为我们需要的节点，否则只取返回2的节点，如此便保证了该方法的正确性。对这种实现还有问题的在下面评论吧。