二叉树习题总结（一）

144. 二叉树的前序遍历

借助栈

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()) {
            TreeNode node = stack.pop();
            res.add(node.val);
            if (node.right != null) stack.push(node.right);
            if (node.left != null) stack.push(node.left);
        }
        return res;
    }
}

递归

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
    List<Integer> res = new ArrayList<>();
    preorderhelper(root, res);
    return res;
    }
    public void preorderhelper(TreeNode node, List<Integer> res) {
        if (node == null)
        return;
        res.add(node.val);
        preorderhelper(node.left, res);
        preorderhelper(node.right, res);
    }
}

94. 二叉树的中序遍历

借助栈

栈先进后出，一直左走直到左子树为空，打印栈内节点，右子树重复

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        while (!stack.isEmpty() || root != null) {
            if (root != null) {
                stack.push(root);
                root = root.left;
            } else {
            //这里可以new一个新节点，也可以使用root
                root = stack.pop();
                res.add(root.val);
                root = root.right;
            }
        }
        return res;
    }
}

递归

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

145. 二叉树的后序遍历

非递归

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()){
            TreeNode node = stack.pop();
            res.add(node.val);
            if (node.left != null)
            stack.push(node.left);
            if (node.right != null) 
            stack.push(node.right);
        }
        Collections.reverse(res);
        return res;
    }
}

递归

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        postorderhelper (root, res);
        return res;
    }
    public void postorderhelper(TreeNode root, List<Integer> res) {
        if (root == null) return;
        postorder(root.left, res);
        postorder(root.right, res);
        res.add(root.val);
    }
}

102. 二叉树的层序遍历

100. 相同的树

简单的递归

class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) return true;
        if (p == null || q == null) return false;
        if (p.val != q.val) return false;
        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    }
}

101. 对称二叉树

//递归
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        return dfs(root.left, root.right);
    }
    public boolean dfs(TreeNode left, TreeNode right) {
        if (left == null && right == null) return true;
        if (left == null || right == null) return false;
        if (left.val != right.val) return false;
        return dfs(left.left, right.right) && dfs(left.right, right.left);
    }
}

用队列实现，先拿出来再比较

class Solution {
    class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null || (root.left == null && root.right == null))
        return true;
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.add(root.left);
        queue.add(root.right);
        while (!queue.isEmpty()) {
            TreeNode left = queue.removeFirst();
            TreeNode right = queue.removeFirst();
            if (left == null && right == null)
            continue;
            if (left == null || right == null)
            return false;
            if (left.val != right.val) 
            return false;
            queue.add(left.left);
            queue.add(right.right);
            queue.add(left.right);
		    queue.add(right.left);
        }
        return true;
    }
}

226. 翻转二叉树

递归

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

借助队列

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        Deque<TreeNode> queue = new ArrayDeque<>();
        queue.add(root);
        while(!queue.isEmpty()) {
            //tmp暂时存放从队列中获取的第一个树节点
            TreeNode tmp = queue.poll();
            //节点left暂时存放当前根节点的左子树
            //进行交换
            TreeNode left = tmp.left;
            tmp.left = tmp.right;
            tmp.right = left;
            if (tmp.left != null) queue.add(tmp.left);
            if (tmp.right != null) queue.add(tmp.right);
        }
        return root;
    }
}