Lowest Common Ancestor of a Binary Search Tree(树中两个结点的最低公共祖先)

题目描述：

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.

　　利用二叉排序树的性质，可以很好地解决这道题。

solution：

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if (root == NULL || p == NULL || q == NULL)return NULL;if (root->val > p->val && root->val > q->val)return lowestCommonAncestor(root->left, p, q);else if (root->val < p->val && root->val < q->val)return lowestCommonAncestor(root->right, p, q);else return root;
}

　　如果只是一颗普通的二叉树呢？

solution：

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if (!root)return NULL;if (root == p || root == q)return root;TreeNode *L = lowestCommonAncestor(root->left, p, q);TreeNode *R = lowestCommonAncestor(root->right, p, q);if (L && R)return root;return L ? L : R;
}

来源：Lowest Common Ancestor of a Binary Tree Part I

如果二叉树的结点存在指向父结点的指针，问题可以转化为求两个单链表的相交结点。

solution：

struct TreeNode {int val;TreeNode *left;TreeNode *right;TreeNode *parent;
};int getHeight(TreeNode *p)
{int height = 0;while (p){height++;p = p->parent;}return height;
}TreeNode* lowestCommonAncestor(TreeNode *p, TreeNode *q)
{int h1 = getHeight(p);int h2 = getHeight(q);if (h1 > h2){swap(h1, h2);swap(p, q);}int dh = h2 - h1;for (int h = 0; h < dh; ++h)q = q->parent;while (p && q){if (p == q)return p;p = p->parent;q = q->parent;}return NULL;
}

来源：Lowest Common Ancestor of a Binary Tree Part II

转载于:https://www.cnblogs.com/gattaca/p/4674664.html

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