LeetCode(五)

Minimum Depth of Binary Tree

public class Solution {
    public int minDepth(TreeNode root) {
        if(root==null) return 0;
        int lh = minDepth(root.left);
        int rh = minDepth(root.right);
        return (lh==0 || rh==0)?lh+rh+1:Math.min(lh,rh)+1;
    }
}

Maxmum Depth of Binary Tree

class Solution {
public:
    int dep = 0;
    int maxDepth(TreeNode* root) {
        dep = dfs(root);
        return dep;
    }
    int dfs(TreeNode* root){
        if(root == NULL) return 0;
        int left = dfs(root->left);
        int right = dfs(root->right);
        return 1+max(left,right);
    }
};

Binary Tree Maximum Path Sum

public class Solution {
    int maxValue;
    public int maxPathSum(TreeNode root) {
        maxValue = Integer.MIN_VALUE;
        maxPathDown(root);
        return maxValue;
    }
    private int maxPathDown(TreeNode node) {
        if (node == null) return 0;
        int left = Math.max(0, maxPathDown(node.left));
        int right = Math.max(0, maxPathDown(node.right));
        maxValue = Math.max(maxValue, left + right + node.val);
        return Math.max(left, right) + node.val;
    }
}

Sum Root to Leaf Numbers

public class Solution {
    public int sumNumbers(TreeNode root) {
        return sum(root, 0);
    }
    public int sum(TreeNode n, int s){
        if (n == null) return 0;
        if (n.right == null && n.left == null) return s*10 + n.val;
        return sum(n.left, s*10 + n.val) + sum(n.right, s*10 + n.val);
    }
}

Binary Tree Paths

public class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> rst = new ArrayList<String>();
        if(root == null) return rst;
        StringBuilder sb = new StringBuilder();
        helper(rst, sb, root);
        return rst;
    }

    public void helper(List<String> rst, StringBuilder sb, TreeNode root){
        if(root == null) return;
        int tmp = sb.length();
        if(root.left == null && root.right == null){
            sb.append(root.val);
            rst.add(sb.toString());
            sb.delete(tmp , sb.length());
            return;
        }
        sb.append(root.val + "->");
        helper(rst, sb, root.left);
        helper(rst, sb, root.right);
        sb.delete(tmp , sb.length());
        return;

    }
}

Binary Tree Right Side View

public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null) return res;
        LinkedList<TreeNode> q = new LinkedList<TreeNode>();
        q.addLast(root);
        while(!q.isEmpty()){
            int tmp = q.size();
            TreeNode tmpNode ;
            for(int i=1;i<=tmp;i++){
                tmpNode = q.getFirst();
                q.removeFirst();
                if(i == tmp){
                    res.add(tmpNode.val);
                }
                if(tmpNode.left!=null) q.addLast(tmpNode.left);
                if(tmpNode.right!=null) q.addLast(tmpNode.right);
            }
        }
        return res;
    }
}

Path Sum I

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        if (sum - root.val == 0 && root.left == null && root.right == null) {
            return true;
        }
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

Path Sum II

public class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>>  res = new ArrayList<List<Integer>>();
        List<Integer> cur = new ArrayList<Integer>();
        help(root,sum,0,res,cur);
        return res;
     }
     public void help(TreeNode p, int sum,int curSum,List<List<Integer>>  res,List<Integer> curList){
         if(p==null) return;
         curList.add(p.val);
         if(curSum+p.val == sum && p.left==null && p.right==null){
             res.add(new ArrayList(curList));
             curList.remove(curList.size()-1);
             return;
         }else{
             help(p.left,sum,curSum+p.val,res,curList);
             help(p.right,sum,curSum+p.val,res,curList);
         }
         curList.remove(curList.size()-1);
         return;
     }
}

Kth Smallest Element in a BST

public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        LinkedList<TreeNode> s = new LinkedList<TreeNode>();
        int cnt = 0;
        TreeNode p = root;
        while(!s.isEmpty() || p!=null){
            if(p!=null){
                s.addFirst(p);
                p = p.left;
            }else{
                p = s.getFirst();
                s.removeFirst();
                cnt++;
                if(cnt==k){return p.val;}
                p = p.right;
            }
        }
        return 0;
    }
}

Unique Binary Search Trees

public class Solution {
    public int numTrees(int n) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        map.put(0,1);
        map.put(1,1);
        return numTrees(n, map);
    }

    private int numTrees(int n, Map<Integer, Integer> map){
        // check memory
        if(map.containsKey(n)) return map.get(n);
        // recursion
        int sum = 0;
        for(int i = 1;i <= n;i++)
            sum += numTrees(i-1, map) * numTrees(n-i, map);
        map.put(n, sum);
        return sum;
    }
}

Unique Binary Search Trees II

public class Solution {
    public List<TreeNode> generateTrees(int n) {
        // DP, compute and store trees for intermediate ranges from 1 to n
        // time: exponential, since for each range the number of trees is combinatorial
        // space: O(n^2)
        if (n == 0) {
            List<TreeNode> result = new ArrayList<TreeNode>();
            //result.add(null);
            return result;
        }
        List<TreeNode>[][] treesForRange = (ArrayList<TreeNode>[][]) new ArrayList[n+2][n+1];
        for (int step = 0; step < n; step++) {
            for (int i = 1; i <= n - step; i++) {
                int j = i + step;
                treesForRange[i][j] = new ArrayList<TreeNode>();
                if (step == 0) {
                    treesForRange[i][j].add(new TreeNode(i));
                    continue;
                }
                // k is a possible root for range i-j
                for (int k = i; k <= j; k++) {
                    // two ugly if blocks for boundary cases:
                    // when k == i, the left subtree is null
                    // when k == j, the right subtree is null
                    if (treesForRange[i][k-1] == null) {
                        treesForRange[i][k-1] = new ArrayList<TreeNode>();
                        treesForRange[i][k-1].add(null);
                    }
                    if (treesForRange[k+1][j] == null) {
                        treesForRange[k+1][j] = new ArrayList<TreeNode>();
                        treesForRange[k+1][j].add(null);
                    }
                    for (TreeNode leftNode : treesForRange[i][k-1]) {
                        for (TreeNode rightNode : treesForRange[k+1][j]) {
                            TreeNode root = new TreeNode(k);
                            root.left = leftNode;
                            root.right = rightNode;
                            treesForRange[i][j].add(root);
                        }
                    }
                }
            }
        }
        return treesForRange[1][n];
    }
}

​