LCA
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/
https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
https://leetcode.com/problems/validate-binary-search-tree/
普通二叉树：
自下向上的解法
我觉得应该再加一层，先检查p，q是不是都在树上。
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(!root) return NULL;
if(root == p || root == q) return root;
TreeNode * left = lowestCommonAncestor(root->left, p, q);
TreeNode * right = lowestCommonAncestor(root->right, p, q);
if(left && right) return root;
if(!left && right) return right;
if(left && !right) return left;
return NULL;
}
};
以上方法无法在不存在的时候返回空。
自上向下的解法，平衡情况 O(n)，最坏情况O(n^2)：
但其实这个也没有检查是否是否两个节点都出现在树上。
http://articles.leetcode.com/lowest-common-ancestor-of-a-binary-tree-part-i/
// Return #nodes that matches P or Q in the subtree.
int countMatchesPQ(Node *root, Node *p, Node *q) {
if (!root) return 0;
int matches = countMatchesPQ(root->left, p, q) + countMatchesPQ(root->right, p, q);
if (root == p || root == q)
return 1 + matches;
else
return matches;
}
Node *LCA(Node *root, Node *p, Node *q) {
if (!root || !p || !q) return NULL;
if (root == p || root == q) return root;
int totalMatches = countMatchesPQ(root->left, p, q);
if (totalMatches == 1)
return root;
else if (totalMatches == 2)
return LCA(root->left, p, q);
else /* totalMatches == 0 */
return LCA(root->right, p, q);
}
非递归方法，注意最后路径分叉和包含问题。
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* t1, TreeNode* t2) {
if(!root || !t1 || !t2) return NULL;
int num = 0;
vector st;
TreeNode * p = root;
TreeNode * pre = NULL;
vector path1, path2;
while(!st.empty() || p!=NULL) {
if(p!=NULL) {
st.push_back(p);
if(p == t1) {
path1 = st;
}
if(p == t2) {
path2 = st;
}
p = p->left;
} else {
TreeNode * n = st.back();
if(n->right != pre && n->right != NULL){
p = n->right;
} else {
st.pop_back();
pre = n;
}
}
}
int m = min(path1.size(), path2.size());
int i;
for(i = 0; i < m; i++) {
if(path1[i] != path2[i] && i > 0) {
return path1[i-1];
}
}
if(i == m) {
return path1[i-1];
}
return NULL;
}
};
BST的相似题目：
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(p->val > q->val) {
swap(p,q);
}
if(root->val >= p->val && root->val <= q->val) {
return root;
}
if(root->val > q->val) {
return lowestCommonAncestor(root->left,p,q);
}
if(root->val < p->val) {
return lowestCommonAncestor(root->right,p,q);
}
}
};
iterative:
https://leetcode.com/discuss/45511/easy-c-recursive-and-iterative-solutions
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
TreeNode* cur = root;
while (true) {
if (p -> val < cur -> val && q -> val < cur -> val)
cur = cur -> left;
else if (p -> val > cur -> val && q -> val > cur -> val)
cur = cur -> right;
else return cur;
}
}
};
把Binary Tree拉成串
Flatten Binary Tree to Linked List
class Solution {
TreeNode * helper(TreeNode * root) {
if(root == NULL) return NULL;
if(!root->left && !root->right) { return root;}
TreeNode * ltail = helper(root->left);
TreeNode * rhead = root->right;
if(root->left) {
root->right = root->left;
root->left = NULL;
ltail->right = rhead;
}
TreeNode * rtail = helper(root->right);
return rtail;
}
public:
void flatten(TreeNode* root) {
if(root == NULL) return;
helper(root);
}
};
106.Construct Binary Tree from Inorder and Postorder Traversal
class Solution {
TreeNode * helper(vector& inorder, vector& postorder, int ii, int ij, int pi, int pj) {
if(ii > ij || pi > pj) {
return NULL;
}
TreeNode * root = new TreeNode(postorder[pj]);
int mid;
for(int i = ii; i <= ij; i++) {
if(inorder[i] == postorder[pj]) {
mid = i;
break;
}
}
int leftlen = mid - ii;
int rightlen = ij - mid;
TreeNode * l = helper(inorder, postorder, ii, mid - 1, pi, pi+ leftlen - 1);
TreeNode * r = helper(inorder, postorder, mid + 1, ij, pi + leftlen, pj - 1);
root->left = l;
root->right = r;
return root;
}
public:
TreeNode* buildTree(vector& inorder, vector& postorder) {
int n = inorder.size();
return helper(inorder,postorder, 0, n-1, 0, n-1);
}
};
98.Validate Binary Search Tree
class Solution {
bool helper(TreeNode * root, TreeNode * smallest, TreeNode * largest) {
if(root == NULL) {return true;}
if(smallest) {
if(root->val <= smallest->val) {return false;}
}
if(largest) {
if(root->val >= largest->val) {return false;}
}
return helper(root->left,smallest,root) && helper(root->right, root, largest);
}
public:
bool isValidBST(TreeNode* root) {
return helper(root,NULL,NULL);
}
};
inorder遍历
https://leetcode.com/discuss/14886/order-traversal-please-rely-buggy-int_max-int_min-solutions
class Solution {
public:
bool isValidBST(TreeNode* root) {
TreeNode* prev = NULL;
return validate(root, prev);
}
bool validate(TreeNode* node, TreeNode* &prev) {
if (node == NULL) return true;
if (!validate(node->left, prev)) return false;
if (prev != NULL && prev->val >= node->val) return false;
prev = node;
return validate(node->right, prev);
}
};