二叉树前中后序遍历（递归+迭代）与层序遍历

1 前序遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */

 // 递归
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        order(root, res);
        return res;
    }

    private void order(TreeNode root, List<Integer> res) {
        if (root == null) return;
        res.add(root.val);
        order(root.left, res);
        order(root.right, res);
    }
}

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

2 后序遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */

 // 递归
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        order(root, res);
        return res;
    }

    private void order(TreeNode root, List<Integer> res) {
        if (root == null) return;
        order(root.left, res);
        order(root.right, res);
        res.add(root.val);
    }
}

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

3 中序遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */

 // 递归
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        order(root, res);
        return res;
    }

    private void order(TreeNode root, List<Integer> res) {
        if (root == null) return;
        order(root.left, res);
        res.add(root.val);
        order(root.right, res);
    }
}

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

4 层序遍历

class Solution {
    public List<Integer> sequenceOrder(TreeNode root) {
        List<Integer> res = new LinkedList<Integer>();
        if (root == null) return res;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int cnt = queue.size();
            for (int i = 0; i < cnt; i++) {
                TreeNode node = queue.poll();
                res.add(node.val);
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
        }
        return res;
    }
}