几道有关二叉树 DFS 和 BFS 的算法题
2019-05-23 本文已影响0人
DejavuMoments
102. Binary Tree Level Order Traversal
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
/**
* 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:
vector<vector<int> > levelOrder(TreeNode* root) {
//return BFS(root);
vector<vector<int> > ans;
DFS(root, 0, ans);
return ans;
}
private:
vector<vector<int> > BFS(TreeNode* root){
// 如果根节点为空，返回空数组
if(!root) return {};
vector<vector<int> > ans;
vector<TreeNode*> cur, next;
cur.push_back(root);
while(!cur.empty()){
// C++ 中 vector.back() 返回当前 vector 最末一个元素的引用
ans.push_back({});
for(TreeNode* node : cur){
// C++ 中 vector.back() 返回当前 vector 最末一个元素的引用
ans.back().push_back(node->val);
if(node->left) next.push_back(node->left);
if(node->right) next.push_back(node->right);
}
cur.swap(next);
next.clear();
}
return ans;
}
void DFS(TreeNode* root, int depth, vector<vector<int>>& ans){
// 如果根节点为空，那么就直接 return ，不做任何处理
// if(!root) return;
if (root == NULL) return;
// works with pre/in/post order
while(ans.size() <= depth) ans.push_back({});
DFS(root->left, depth+1, ans);
DFS(root->right, depth+1, ans);
ans[depth].push_back(root->val); // post-order
}
};
103. Binary Tree Zigzag Level Order Traversal
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
/**
* 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:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> ans;
if(root == nullptr) return ans;
queue<TreeNode*> q{{root}};
bool left2right = true;
while(!q.empty()){
int level = 0;
vector<int> level_nodes;
for(int i = 0, n = q.size(); i < n; i++){
auto node = q.front();
q.pop();
if(left2right){
level_nodes.push_back(node->val);
}else{
level_nodes.insert(level_nodes.begin(), node->val);
}
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
// after this level
left2right=!left2right;
ans.push_back(level_nodes);
}
return ans;
}
};
107. Binary Tree Level Order Traversal II
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
/**
* 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:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> ans;
if(root == nullptr) return ans;
queue<TreeNode*> q{{root}};
while(!q.empty()){
vector<int> level_nodes;
for(int i = 0, n = q.size(); i < n; i++){
auto node = q.front();
q.pop();
level_nodes.push_back(node->val);
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
ans.insert(ans.begin(),level_nodes);
}
return ans;
}
};
104. Maximum Depth of Binary Tree
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Note: A leaf is a node with no children.
Example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its depth = 3.
Solution 1 DFS
/**
* 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:
int maxDepth(TreeNode* root) {
if(root == nullptr) return 0;
// 以下两行可以加快速度，从 16ms 到 8ms
if(!root->left) return 1 + maxDepth(root->right);
if(!root->right) return 1 + maxDepth(root->left);
return 1 + max(maxDepth(root->right), maxDepth(root->left));
}
};
Solution 2 BFS
4ms
/**
* 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:
int maxDepth(TreeNode* root) {
if(root == nullptr) return 0;
queue<TreeNode*> q{{root}};
int max_depth = 0;
while(!q.empty()){
max_depth++;
for(int i = 0, n = q.size(); i < n; i++){
auto node = q.front();
q.pop();
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
}
return max_depth;
}
};
111. Minimum Depth of Binary Tree
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. Note: A leaf is a node with no children.
Example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its minimum depth = 2.
Solution 1 : DFS
/**
* 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:
int minDepth(TreeNode* root) {
if(root == nullptr) return 0;
if(root->left == nullptr) return minDepth(root->right) + 1;
if(root->right == nullptr) return minDepth(root->left) + 1;
return 1 + min(minDepth(root->right), minDepth(root->left));
}
};
Solution 2 : BFS
BFS reaches the minimal depth leaf node before DFS.
class Solution {
public:
int minDepth(TreeNode* root) {
if(root == nullptr) return 0;
int min_depth = 0;
queue<TreeNode*> q {{root}};
while(!q.empty()){
min_depth++;
for(int i = q.size(); i > 0; i--){
auto node = q.front();
q.pop();
if(!node->left && !node->right) return min_depth;
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
}
return min_depth;
}
};
513. Find Bottom Left Tree Value
Given a binary tree, find the leftmost value in the last row of the tree.
Example 1:
Input:
2
/ \
1 3
Output:
1
Example 2:
Input:
1
/ \
2 3
/ / \
4 5 6
/
7
Output:
7
Note: You may assume the tree (i.e., the given root node) is not NULL.
/**
* 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:
int findBottomLeftValue(TreeNode* root) {
int ans = 0;
if(root == nullptr) return 0;
queue<TreeNode*> q{{root}};
while(!q.empty()){
vector<int> level_nodes;
for(int i = 0, n = q.size(); i < n; i++){
TreeNode* node = q.front();
q.pop();
level_nodes.push_back(node->val);
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
ans = level_nodes[0];
}
return ans;
}
};
