面试二叉树题目汇总

二叉树的最大距离（即相距最远的两个叶子节点，编程之美，百度面试题）

int HeightOfBinaryTree(BinaryTreeNode *pNode, int &nMaxDistance){
	if (pNode == NULL)
		return -1;   //空节点的高度为-1
	//递归
	int nHeightOfLeftTree = HeightOfBinaryTree(pNode->m_pLeft, nMaxDistance) + 1;   //左子树的的高度加1
	int nHeightOfRightTree = HeightOfBinaryTree(pNode->m_pRight, nMaxDistance) + 1;   //右子树的高度加1
	int nDistance = nHeightOfLeftTree + nHeightOfRightTree;    //距离等于左子树的高度加上右子树的高度+2
	nMaxDistance = nMaxDistance > nDistance ? nMaxDistance : nDistance;            //得到距离的最大值
	return max(nHeightOfLeftTree, nHeightOfRightTree);
}

二叉树的深度

int TreeDepth(TreeNode *pRoot){
	if(!pRoot) return 0;
	int left = TreeDepth(pRoot->left);
	int right = TreeDepth(pRoot->right);
	return max(left, right)+1;
}

重建二叉树

对称二叉树(leetcode 101)

bool isSymmetric(TreeNode* root) {
	if(!root) return true;
    return isSymmetric(root->left, root->right);
}
bool isSymmetric(TreeNode *left, TreeNode *right){
    if(!left && !right) return true;
    if(!left || !right || left->val!=right->val) return false;

    return isSymmetric(left->left, right->right) && 
    	isSymmetric(left->right, right->left);
}

二叉树的下一个节点

树的子结构

bool isSubtree(TreeNode* s, TreeNode* t) {
    if (!s) return false;
    if (isSame(s, t)) return true;
    return isSubtree(s->left, t) || isSubtree(s->right, t);
}
bool isSame(TreeNode* s, TreeNode* t) {
    if (!s && !t) return true;
    if (!s || !t) return false;
    if (s->val != t->val) return false;
    return isSame(s->left, t->left) && isSame(s->right, t->right);
}

二叉搜索树的后序遍历序列

二叉树中和为某一值的路径

二叉搜索树与双向链表

序列化二叉树

树中两个节点的最低公共祖先

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if(!root || root==p || root==q) return root;
    
    TreeNode *left = lowestCommonAncestor(root->left, p, q);
    if(left && left!=p && left!=q) return left;
    TreeNode *right = l
}

二叉树的非递归遍历

前序遍历(leetcode 144)

vector<int> preorderTraversal(TreeNode* root) {
	if(!root) return {};
	
	stack<TreeNode*> s;
	vector<int> res;
	s.push(root);
	while(!s.empty()){
		TreeNode *p = s.top();
		s.pop();
		res.push_back(p->val);
		if(p->right) s.push(p->right);
		if(p->left) s.push(p->left);
	}
	return res; 
}

中序遍历(leetcode 94)

vector<int> inorderTraversal(TreeNode *root) {
	vector<int> res;
	stack<TreeNode*> s;
	TreeNode *p = root;
	
	while (p || !s.empty()) {
		while (p) {
			s.push(p);
			p = p->left;
		}
		p = s.top(); s.pop();
		res.push_back(p->val);
		p = p->right;
	}
	return res;
}

后序遍历(leetcode 145)

vector<int> postorderTraversal(TreeNode* root) {
    if(!root) return {};
    stack<TreeNode*> s;
    vector<int> res;
    s.push(root);
    
    while(!s.empty()){
        TreeNode *cur = s.top(); s.pop();
        if(cur->left) s.push(cur->left);
        if(cur->right) s.push(cur->right);
        res.insert(res.begin(), cur->val);
    }
    return res;
}

平衡二叉树(leetcode 110)

bool isBalanced(TreeNode* root) {
    int depth = 0;
    return isBalanced(root, depth);
}
bool isBalanced(TreeNode *root, int &depth){
    if(!root) return true;
    int left=0, right=0;
    bool leftBalanced = isBalanced(root->left, left);
    bool rightBalanced = isBalanced(root->right, right);
    depth = max(left, right) + 1;
    if(!leftBalanced || !rightBalanced) return false;
    if(abs(left-right)>1) return false;
    return true;
}

未完待续, 有时间再补充