hot100-二叉树

二叉树

二叉树递归

相当于这个的顺序来回调换

class Solution {
    private List<Integer> res = new ArrayList<>();
    public List<Integer> inorderTraversal(TreeNode root) {
        if(root == null)return res;
        inorderTraversal(root.left);
        res.add(root.val);
        inorderTraversal(root.right);
        return res;
    }
}

二叉树迭代

前序遍历

迭代法

public List<Integer> preOrderTraversal(TreeNode root) {
    List<Integer> res = new ArrayList<Integer>();
    if (root == null) {
        return res;
    }
    
    Deque<TreeNode> stack = new LinkedList<TreeNode>();
    TreeNode node = root;
    while (!stack.isEmpty() || node != null) {
        while (node != null) {
            res.add(node.val);
            stack.push(node);
            node = node.left;
        }
        node = stack.pop();
        node = node.right;
    }
    return res;
}

中序遍历

迭代法

public List<Integer> preOrderTraversal(TreeNode root) {
    List<Integer> res = new ArrayList<Integer>();
    if (root == null) {
        return res;
    }
    
    Deque<TreeNode> stack = new LinkedList<TreeNode>();
    TreeNode node = root;
    while (!stack.isEmpty() || node != null) {
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
        node = stack.pop();
        res.add(node.val);
        node = node.right;
    }
    return res;
}

染色法

缺点是要写一个pair的类，优点是只需要更改顺序就可以使三个顺序都能写

class Pair<K, V> {
    private K key;
    private V value;

    public Pair(K key, V value) {
        this.key = key;
        this.value = value;
    }

    public K getKey() {
        return key;
    }

    public V getValue() {
        return value;
    }
}

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        Deque<Pair<Integer, TreeNode>> stack = new ArrayDeque<>();
        stack.push(new Pair<>(0,root));
        while(!stack.isEmpty()){
            Pair<Integer,TreeNode> newPair = stack.pop();
            int color = newPair.getKey();
            TreeNode node = newPair.getValue();
            if(node == null)continue;
            if(color == 0){
                stack.push(new Pair<>(0,node.right));
                stack.push(new Pair<>(1,node));
                stack.push(new Pair<>(0,node.left));
            }else{
                res.add(node.val);
            }
        }
        return res;
    }
}

Morris法

①第一个循环：是否为空

②判断左子树是否为空，是则记录+进入右子树，不是则进入左子树

③如果最右的最右为空，则链接，进入左子树。如果最右的最右为root，则断联，记录，进入右子树。

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        while(root !=null){
            if(root.left == null){
                res.add(root.val);
                root = root.right;
            }else{
                //有左子树的情况
                TreeNode pre = root.left;
                while(pre.right != null && pre.right != root){
                    pre = pre.right;
                }
                if(pre.right == null){
                    pre.right = root;
                    root = root.left;
                }else{
                    pre.right = null;//断开
                    res.add(root.val);
                    root = root.right;
                }
            }
        }
        return res;
    }
}

后序遍历

反转法

其实就是将递归暗处的栈变成明面

public class PostorderTraversal {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;

        Stack<TreeNode> stack = new Stack<>();
        Stack<Integer> output = new Stack<>();
        stack.push(root);

        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            output.push(node.val);

            if (node.left != null) stack.push(node.left);
            if (node.right != null) stack.push(node.right);
        }

        while (!output.isEmpty()) {
            result.add(output.pop());
        }

        return result;
    }

访问标记法（染色法）

（使用额外的标记来指示节点是否已经访问）

public class PostorderTraversal {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;

        Stack<TreeNode> stack = new Stack<>();
        TreeNode prev = null;

        while (!stack.isEmpty() || root != null) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }

            root = stack.peek();
            if (root.right == null || root.right == prev) {
                result.add(root.val);
                stack.pop();
                prev = root;
                root = null; // We have finished this node
            } else {
                root = root.right; // Move to the right child
            }
        }

        return result;
    }

Morris法（线性时间，常数空间）

Morris 遍历法通过在遍历过程中使用指针而避免了使用栈或递归，从而节省空间。

public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;

        TreeNode dummy = new TreeNode(0);
        dummy.left = root;
        TreeNode curr = dummy;

        while (curr != null) {
            if (curr.left == null) {
                curr = curr.right;
            } else {
                TreeNode prev = curr.left;
                while (prev.right != null && prev.right != curr) {
                    prev = prev.right;
                }

                if (prev.right == null) {
                    prev.right = curr;
                    curr = curr.left;
                } else {
                    reverse(curr.left, prev);
                    TreeNode temp = prev;
                    while (temp != null) {
                        result.add(temp.val);
                        temp = temp.right;
                    }
                    reverse(prev, curr.left);
                    prev.right = null;
                    curr = curr.right;
                }
            }
        }

        return result;
    }

    private void reverse(TreeNode from, TreeNode to) {
        if (from == to) return;
        TreeNode x = from, y = from.right;
        while (x != to) {
            x.right = y.right;
            y.right = x;
            x = y;
            y = y.right;
        }
    }

104. 二叉树的最大深度

解法一、递归

class Solution {
    public int maxDepth(TreeNode root) {
        if(root == null)return 0;
        int leftMax = maxDepth(root.left);
        int rightMax = maxDepth(root.right);
        return Math.max(leftMax,rightMax) + 1;
    }
}

226. 翻转二叉树

解法一、递归

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null)return root;
        invertTree(root.left);
        invertTree(root.right);
        TreeNode tmp = root.left;
        root.left = root.right;
        root.right = tmp;
        return root;
    }
}

101. 对称二叉树

解法一、递归

class Solution {
    public boolean isSymmetric(TreeNode root) {
        return isS(root.left,root.right);
    }
    private boolean isS(TreeNode left,TreeNode right){
        if(left == null || right == null)return left == right;
        return left.val == right.val && isS(left.left,right.right)&&isS(left.right,right.left);
    }
}

解法二、迭代

因为List情况下不能add null，所以改换成Queue。不过不改也可以，只需要在null的情况下构建新节点，总之就是改换边界条件

class Solution {
    public boolean isSymmetric(TreeNode root) {
        return isS(root,root);
    }
    private boolean isS(TreeNode left,TreeNode right){
        Queue<TreeNode> res = new LinkedList<>();
        res.add(left);
        res.add(right);
        while (!res.isEmpty()){
            TreeNode u = res.poll();
            TreeNode v = res.poll();
            if (u == null && v == null) {
                continue;
            }
            if ((u == null || v == null) || (u.val != v.val)) {
                return false;
            }
            res.offer(u.left);
            res.offer(v.right);
            res.offer(u.right);
            res.offer(v.left);
        }
        return true;
    }
}

543. 二叉树的直径

解法一、递归

class Solution {
    private int res = 0;
    public int diameterOfBinaryTree(TreeNode root) {
        dfs(root);
        return res;
    }
    private  int dfs(TreeNode root){
        if(root == null)return -1;
        int l = dfs(root.left)+1;
        int r = dfs(root.right)+1;
        res = Math.max(res,l+r);
        return Math.max(l,r);
    }
}

102. 二叉树的层序遍历

解法一层序遍历

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<>();
        List<List<Integer>> res = new LinkedList<>();
        q.offer(root);
        while(!q.isEmpty()){
            Queue<TreeNode> p = q;
            q = new LinkedList<>();
            List<Integer> tmp = new LinkedList<>();
            while(!p.isEmpty()){
                TreeNode node = p.poll();
                if(node == null)continue;
                q.add(node.left);
                q.add(node.right);
                tmp.add(node.val);
            }
            if(!tmp.isEmpty())res.add(tmp);
        }
        return res;
    }
}

108. 将有序数组转换为二叉搜索树

解法一、递归

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        int n = nums.length,mid = n/2;
        return create(nums,0,n);
    }
    private TreeNode create(int[] nums,int x,int y){
        if(x==y)return null;
        int mid = x + (y-x)/2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = create(nums,x,mid);
        node.right = create(nums,mid+1,y);
        return node;
    }
}

98. 验证二叉搜索树

解法一、递归

class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root == null)return true;
        return BST(root.left,Long.MIN_VALUE,root.val) && BST(root.right,root.val,Long.MAX_VALUE);
    }
    public boolean BST(TreeNode root,long x,long y) {
        if(root == null)return true;
        if(root.val <= x || root.val >= y )return false;
        return BST(root.left,x,root.val) && BST(root.right,root.val,y);
    }
}

解法二、中序递增

中序出来的数组一定是递增的，同时，递增数组中序构建也一定是BST

class Solution {
    public boolean isValidBST(TreeNode root) {
       List<Integer> tmp = new LinkedList<>();
       while(root != null){
           if(root.left == null){
               tmp.add(root.val);
               root = root.right;
           }else{
               TreeNode node = root.left;
               while(node.right != null && node.right != root){
                   node = node.right;
               }
               if(node.right == null){
                   node.right = root;
                   root = root.left;
               }else{
                   node.right = null;
                   tmp.add(root.val);
                   root = root.right;
               }
           }
       }
       int n = tmp.size(),num = tmp.get(0);
       for(int i = 1;i < n;i++){
           if(tmp.get(i) <= num)return false;
           num = tmp.get(i);
       }
       return true;
    }
}

230. 二叉搜索树中第 K 小的元素

解法一、中序递增

98的变式。在while里判断一下tmp.size()和k的关系剪枝，可以效率提升一半多。

class Solution {
    public int kthSmallest(TreeNode root, int k) {
       List<Integer> tmp = new LinkedList<>();
       while(root != null){
           if(root.left == null){
               tmp.add(root.val);
               root = root.right;
           }else{
               TreeNode node = new TreeNode();
               while(node.right != null && node.right != root){
                   node = node.right;
               }
               if(node.right == null){
                   node.right = root;
               }else{
                   node.right = null;
                   tmp.add(root.val);
                   root = root.right;
               }
           }
       }
       return tmp.get(k);
    }
}

解法二、dfs

第一个if是边界，第二个if是剪枝，第三个是答案赋值判断

class Solution {
    private int k,res= 0;
    public int kthSmallest(TreeNode root, int k) {
        this.k = k;
        dfs(root);
       return res;
    }
    private void dfs(TreeNode root){
        if(root==null)return;
        dfs(root.left);
        if(k==0)return;
        if(--k==0)res = root.val;
        dfs(root.right);
    }
}

199. 二叉树的右视图

解法一、递归

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        Deque<TreeNode> a = new LinkedList<>();
        a.add(root);
        List<Integer> res = new LinkedList<>();
        if(root == null)return res;
        while(!a.isEmpty()){
            Deque<TreeNode> q = a;
            a = new LinkedList<>();
            res.add(q.getLast().val);
            while(!q.isEmpty()){
                TreeNode node = q.pollLast();
                if(node.right != null)a.addFirst(node.right);
                if(node.left != null)a.addFirst(node.left);
            }
        }
        return res;
    }
}

105. 从前序与中序遍历序列构造二叉树

解法一、递归

靠前序确定节点靠中序确定左右树从顶至下递归进行

两个改进：①搜索改进，甚至使用哈希表O（1）。②更改函数传入，l和r，避免复制数组的开销

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        int n = preorder.length;
        if(n==0)return null;
        int index = findNum(inorder,preorder[0]);
        int [] l = Arrays.copyOfRange(inorder,0,index);
        int [] r = Arrays.copyOfRange(inorder,index+1,n);
        int [] pre1 = Arrays.copyOfRange(preorder,1,1+index);
        int [] pre2 = Arrays.copyOfRange(preorder,1+index,n);
        TreeNode left = buildTree(pre1,l);
        TreeNode right = buildTree(pre2,r);
        return new TreeNode(preorder[0],left,right);
    }
    public int findNum(int[] nums,int num){
        int n = nums.length;
        for(int i = 0;i < n;i++){
            if(nums[i] == num)return i;
        }
        return -1;
    }
}

437. 路径总和 III

解法一、递归哈希

务必恢复现场

class Solution {
    private int ans;
    public int pathSum(TreeNode root, int targetSum) {
        Map<Long,Integer>cnt = new HashMap<>();
        cnt.put(0L,1);
        dfs(root,0,targetSum,cnt);
        return ans;
    }
    private void dfs(TreeNode root,long s,int targetSum,Map<Long,Integer> cnt){
        if(root == null)return;
        s+= root.val;
        ans += cnt.getOrDefault(s - targetSum, 0);
        cnt.merge(s,1,Integer::sum);
        dfs(root.left,s,targetSum,cnt);
        dfs(root.right,s,targetSum,cnt);
        cnt.merge(s,-1,Integer::sum);
    }
}

236. 二叉树的最近公共祖先

解法一、递归找路径，判断两条路径共同开头

你根本无法理解这个到底有多慢jpg

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        List<TreeNode> l = new LinkedList<>();
        List<TreeNode> r = new LinkedList<>();
        dfs(root,p,l);
        dfs(root,q,r);
        int n = l.size(),m = r.size(),x = 0,y = 0;
        while(x < n && y < m && l.get(x) == r.get(y)){
            x++;
            y++;
        }
        TreeNode res;
        if(x>=n)res = l.get(n-1);
        else if(y>=m)res = r.get(m-1);
        else res = l.get(x-1);
        return res;
    }
    public boolean dfs(TreeNode root,TreeNode p,List<TreeNode> list){
        if(root == null)return false;
        list.add(root);
        if(root == p){
            return true;
        }
        //都是错的才remove
        if(dfs(root.left,p,list) || dfs(root.right,p,list))return true;
        list.remove(list.size()-1);
        return false;
    }
}

解法二、递归

进行了一个很大的剪枝。都=null即都没查到，两个都非null即当前节点就公共，p在q下面则返回q即可。

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null || root == p || root == q)return root;
        TreeNode left = lowestCommonAncestor(root.left,p,q);
        TreeNode right = lowestCommonAncestor(root.right,p,q);
        if(left != null && right != null)return root;
        return left != null ? left : right;
    }

}

124. 二叉树中的最大路径和

解法一、递归

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