Lowest Common Ancestor of a Binary Tree -- leetcode

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

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).”

        _______3______       /              \    ___5__          ___1__   /      \        /      \   6      _2       0       8         /  \         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

算法一：递归

1. 先在左子树找LCA；再在右子树找LCA。 返回过程中，如果没有找到LCA，找到p, 或者q，也将其返回。

2. 如果左右子树，返回值都不为空，则本节点为LCA。  说明p,q 分别在左，右子树，或者相反。

要注意的事，此题有个陷阱，比较节点，要比较指针值，而不能比较指指向的val字段。比如不能 root->val == p-val ， 而要 root == p。

因为测试用例中，有大量重复的val。也就意味着LCA不唯一了。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        if (!root || root == p || root == q)            return root;                auto left = lowestCommonAncestor(root->left, p, q);        auto right = lowestCommonAncestor(root->right, p, q);        if (left && right)            return root;        else            return left ? left : right;    }};

算法二，先序遍历

在进行先序遍历中，

1.访问本结点和左子树，右子树后，发现p, q值已经找到，则本节点就是LCA

class Solution {public:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        bool flagp = false, flagq = false;        return dfs(root, p, q, flagp, flagq);    }        TreeNode* dfs(TreeNode* root, TreeNode* p, TreeNode* q, bool& flagp, bool& flagq) {        if (!root)            return root;                    auto flagp_ = flagp;        auto flagq_ = flagq;        flagp = root == p ? !flagp : flagp;        flagq = root == q ? !flagq : flagq;        auto left = dfs(root->left, p, q, flagp, flagq);        auto right = dfs(root->right, p, q, flagp, flagq);        if (left || right)            return left ? left : right;        if (flagp_ != flagp && flagq_ != flagq)            return root;        return NULL;    }};

也可以稍作改进一下，如果访问本节点和左子树已经找到LCA时，不再访问右子树了。如下：

class Solution {public:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        bool flagp = false, flagq = false;        return dfs(root, p, q, flagp, flagq);    }        TreeNode* dfs(TreeNode* root, TreeNode* p, TreeNode* q, bool& flagp, bool& flagq) {        if (!root)            return root;                    auto flagp_ = flagp;        auto flagq_ = flagq;        flagp = root == p ? !flagp : flagp;        flagq = root == q ? !flagq : flagq;        auto ans = dfs(root->left, p, q, flagp, flagq);        if (ans)            return ans;        if (flagp_ != flagp && flagq_ != flagq)            return root;                    ans = dfs(root->right, p, q, flagp, flagq);        if (ans)            return ans;        if (flagp_ != flagp && flagq_ != flagq)            return root;        return NULL;    }};