107. Binary Tree Level Order Traversal II

Difficulty: Easy

Frequency: N/A

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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

---

My solution:

Data structure:

Steps:

Complexity:

Runtime:

Space:

Test cases:

Corner cases:

Code:

/**
 * 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>> results;
        if (!root) return results;
        queue<TreeNode*> myQ;
        myQ.push(root);
        while (!myQ.empty()) {
            int q_size = myQ.size();
            vector<int> tmp;
            while (q_size) {
                TreeNode *cur = myQ.front();
                tmp.push_back(cur->val);
                myQ.pop();
                if (cur->left) myQ.push(cur->left);
                if (cur->right) myQ.push(cur->right);
                q_size--;
            }
            results.push_back(tmp);
        }
        return vector<vector<int>> (results.rbegin(), results.rend());
    }
};

---

Another solution:

Data structure:

steps:

Complexity:

Runtime:

Space:

Code:

---

Things to learn:

Advertisements

102. Binary Tree Level Order Traversal

Difficulty: Easy

Frequency: N/A

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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

---

My solution:

Data structure:

Steps:

Complexity:

Runtime:

Space:

Test cases:

Corner cases:

Code:

/**
 * 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) {
        vector<vector<int>> results;
        if (!root) return results;
        queue<TreeNode*> myQ;
        myQ.push(root);
        while (!myQ.empty()) {
            vector<int> tmp;
            int q_size = myQ.size();
            while (q_size) {
                TreeNode *cur = myQ.front();
                tmp.push_back(cur->val);
                myQ.pop();
                if (cur->left) myQ.push(cur->left);
                if (cur->right) myQ.push(cur->right);
                q_size--;
            }
            results.push_back(tmp);
        }
        return results;
    }
};

---

Another solution:

Data structure:

steps:

Complexity:

Runtime:

Space:

Code:

---

Things to learn:

124. Binary Tree Maximum Path Sum

Difficulty: Hard

Frequency: N/A

Given a binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

---

My solution:

Data structure:

Steps:

Complexity:

Runtime:

Space:

Test cases:

Corner cases:

Code:

/**
 * 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 maxPathSum(TreeNode* root) {
        int maxVal = INT_MIN;
        getMaxSum(root, maxVal);
        return maxVal;
    }
    int getMaxSum(TreeNode* node, int& maxVal) {
        if (!node) return 0;
        int l = getMaxSum(node->left, maxVal);
        int r = getMaxSum(node->right, maxVal);
        if (l < 0) l = 0;
        if (r < 0) r = 0;
        if (l + r + node->val > maxVal) maxVal = l + r + node->val;
        return node->val += max(l, r);
    }
};

---

Another solution:

Data structure:

steps:

Complexity:

Runtime:

Space:

Code:

---

Things to learn:

099. Recover Binary Search Tree

Difficulty: Hard

Frequency: N/A

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

---

My solution:

Data structure:

Steps:

Complexity:

Runtime:

Space:

Test cases:

Corner cases:

Code:

/**
 * 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 {
private:
    TreeNode *first = NULL, *second = NULL, *pre = new TreeNode(INT_MIN);
public:
    void recoverTree(TreeNode* root) {
        tranverseTree(root);
        int tmp = first->val;
        first->val = second->val;
        second->val = tmp;
    }
    void tranverseTree(TreeNode* node) {
        if (!node) return;
        tranverseTree(node->left);
        if (!first && node->val < pre->val) first = pre;
        if (first && node->val < pre->val) second = node;
        pre = node;
        tranverseTree(node->right);
    }
};

---

Another solution:

Data structure:

steps:

Complexity:

Runtime:

Space:

Code:

---

Things to learn: