Clone Graph

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

BFS: 用一个HashMap存储已经生成过的点，避免重复生成相同的点。

public class Solution {
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if (node == null) {
            return null;
        }
        
        HashMap<UndirectedGraphNode, UndirectedGraphNode> map = 
        new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
        Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
        queue.offer(node);
        map.put(node, new UndirectedGraphNode(node.label));
        while (!queue.isEmpty()) {
            UndirectedGraphNode cur = queue.poll();
            UndirectedGraphNode newcur = map.get(cur);
            for (UndirectedGraphNode neig : cur.neighbors) {
                //if contains, this node has already been searched
                if (!map.containsKey(neig)) {
                    map.put(neig, new UndirectedGraphNode(neig.label));
                    queue.offer(neig);
                }
                newcur.neighbors.add(map.get(neig));
            }
        }
        return map.get(node);
    }
}

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,#,#,15,7},

    3

   / \

  9  20

     /    \

   15   7

return its zigzag level order traversal as:

[

  [3],

  [20,9],

  [15,7]

]

BFS: 用两个stack来实现层遍历

public class Solution {
    public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if (root == null) {
            return res;
        }
        
        Stack<TreeNode> cur = new Stack<TreeNode>();
        Stack<TreeNode> next = new Stack<TreeNode>();
        cur.push(root);
        boolean reverse = true;
        
        while (!cur.isEmpty()) {
            ArrayList<Integer> level = new ArrayList<Integer>();
            while (!cur.isEmpty()) {
                TreeNode node = cur.pop();
                if (reverse) {
                    if (node.left != null) {
                        next.push(node.left);
                    }
                    if (node.right != null) {
                        next.push(node.right);
                    }
                } else {
                    if (node.right != null) {
                        next.push(node.right);
                    }
                    if (node.left != null) {
                        next.push(node.left);
                    }
                }
                level.add(node.val);
            }
            reverse = !reverse;
            Stack<TreeNode> temp = cur;
            cur = next;
            next = temp;
            res.add(level);
        }
        return res;
    }
}

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.

DFS:

public class Solution {
    //Time: O(n)
    int min = Integer.MAX_VALUE;
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        getMin(root, 1);
        return min;
    }
    
    private void getMin(TreeNode root, int depth) {
        if (root.left == null && root.right == null) {
            min = Math.min(min, depth);
        }
        
        if (root.left != null) {
            getMin(root.left, depth + 1);
        }
        if (root.right != null) {
            getMin(root.right, depth + 1);
        }
    }
}

BFS:

public class Solution {
    //Time: O(n)
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        int length = 1;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
                if (cur.left == null && cur.right == null) {
                    return length;
                }
            }
            length++;
        }
        return length;
    }
}

Sum Root to Leaf Numbers

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1

   /  \

  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

DFS: 

public class Solution {
    //Time: O(n)
    private int sum = 0;
    public int sumNumbers(TreeNode root) {
        if (root == null) {
            return sum;
        }
        getSum(root, 0);
        return sum;
    }
    
    private void getSum(TreeNode root, int cur) {
        cur = cur * 10 + root.val;
        if (root.left == null && root.right == null) {
            sum += cur;
            return;
        }
        if (root.left != null) {
            getSum(root.left, cur);
        }
        if (root.right != null) {
            getSum(root.right, cur);
        }
    }
}

BFS: 

public class Solution {
    //Time: O(n)
    public int sumNumbers(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        int sum = 0;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        Queue<Integer> queuesum = new LinkedList<Integer>();
        queue.offer(root);
        queuesum.offer(root.val);
        
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            int cursum = queuesum.poll();
            if (cur.left != null) {
                queue.offer(cur.left);
                queuesum.offer(cursum * 10 + cur.left.val);
            }
            if (cur.right != null) {
                queue.offer(cur.right);
                queuesum.offer(cursum * 10 + cur.right.val);
            }
            if (cur.left == null && cur.right == null) {
                sum += cursum;
            }
        }
        return sum;
    }
}

Path Sum I && II

I: Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

DFS: 

public class Solution {
    //Time: O(n)
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        return checkPath(root, root.val, sum);
    }
    
    private boolean checkPath(TreeNode root, int cur, int sum) {
        if (root.left == null && root.right == null) {
            if (cur == sum) {
                return true;
            }
            return false;
        }
        
        boolean left = false;
        boolean right = false;;
        if (root.left != null) {
            left = checkPath(root.left, cur + root.left.val, sum);
        }
        if (root.right != null) {
            right = checkPath(root.right, cur + root.right.val, sum);
        }
        return left || right;
    }
}

BFS: 

public class Solution {
    //Time: O(n)
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        Queue<Integer> queueSum = new LinkedList<Integer>();
        queue.offer(root);
        queueSum.offer(root.val);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            int cursum = queueSum.poll();
            if (cur.left != null) {
                queue.offer(cur.left);
                queueSum.offer(cursum + cur.left.val);
            }
            if (cur.right != null) {
                queue.offer(cur.right);
                queueSum.offer(cursum + cur.right.val);
            }
            if (cur.left == null && cur.right == null && cursum == sum) {
                return true;
            }
        }
        return false;
    }
}

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

DFS: 

public class Solution {
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if (root == null) {
            return res;
        }
        ArrayList<Integer> tmp = new ArrayList<Integer>();
        path(res, tmp, root, sum);
        return res;
    }
    
    private void path(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> tmp,
                      TreeNode root, int sum) {
        if (root.left == null && root.right == null) {
            tmp.add(root.val);
            if (sum == root.val) {
                res.add(new ArrayList(tmp));
            }
            tmp.remove(tmp.size() - 1);
            return;
        }   
        
        tmp.add(root.val);
        if (root.left != null) {
            path(res, tmp, root.left, sum - root.val);
        }
        if (root.right != null) {
            path(res, tmp, root.right, sum - root.val);
        }
        tmp.remove(tmp.size() - 1);
    }
}

Binary Tree Level Order Traversal I && II

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]
]

用一个queue进行BFS。第二个只需把25行改成add(0, level)

public class Solution {
    //Time: O(n)  Space: O(n)
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if (root == null) {
            return res;
        }
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            ArrayList<Integer> level = new ArrayList<Integer>();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                level.add(cur.val);
                
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            res.add(level);
        }
        return res;
    }
}