2.二叉树的遍历（前中后迭代+递归）+层序+通用模板力扣版

2.二叉树的遍历

2.1二叉树的深度优先遍历

2.1.1递归

2.1.1.1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;
 * }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList();
        pre(root, res);
        return res;

    }

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

    }
}

2.1.1.2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;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res=new ArrayList();
        ind(root,res);
        return res;
    }
    public void ind(TreeNode root,List<Integer> res){
        if(root==null){
            return;
        }
        ind(root.left,res);
        res.add(root.val);
        ind(root.right,res);
    }
}

2.1.1.3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;
 *     }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res=new ArrayList();
        post(root,res);
        return res;

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

2.1.2迭代

2.1.2.1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;
 *     }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        LinkedList<TreeNode> stack=new LinkedList();
        List<Integer> res=new ArrayList();
        TreeNode cur=root;
        while(cur!=null||!stack.isEmpty()){
            if(cur!=null){
                stack.push(cur);
                res.add(cur.val);
                cur=cur.left;
            }
            else{
                TreeNode pop=stack.pop();
                cur=pop.right;
            }
        }
        return res;
    }
}

2.1.2.2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;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res=new ArrayList();
        TreeNode cur=root;
        LinkedList<TreeNode> stack=new LinkedList();
        while(cur!=null||!stack.isEmpty()){
            if(cur!=null){
                stack.push(cur);
                cur=cur.left;
            }
            else{
                TreeNode pop=stack.pop();
                res.add(pop.val);
                cur=pop.right;
            }
        }
        return res;
    }
}

2.1.2.3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;
 *     }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res=new ArrayList();
        LinkedList<TreeNode> stack=new LinkedList();
        TreeNode cur=root;
        TreeNode pop=null;
        while(cur!=null||!stack.isEmpty()){
            if(cur!=null){
                stack.push(cur);
                cur=cur.left;
            }
            else{
                TreeNode peek=stack.peek();
                if(peek.right==null||peek.right==pop){
                    pop=stack.pop();
                    res.add(pop.val);
                }
                else{
                    cur=peek.right;
                }
            }
        }
        return res;
    }

}

2.1.2.4前中后通用模板

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res=new ArrayList();
        LinkedList<TreeNode> stack=new LinkedList();
        TreeNode cur=root;
        TreeNode pop=null;
        while(cur!=null||!stack.isEmpty()){
            // 待处理左子树
            if(cur!=null){
                stack.push(cur);
             // 前序遍历   res.add(cur.val);
                cur=cur.left;
            }
            else{
                TreeNode peek=stack.peek();
                //没有右子树
                if(peek.right==null){
                    res.add(peek.val);
                    pop=stack.pop();
                    // 后续遍历 res.add(pop.val);
                    
                }
                // 右子树处理完
                else if(peek.right==pop){
                    pop=stack.pop();
                    // 后续遍历res.add(pop.val);
                }
                // 待处理右子树
                else{
                    res.add(peek.val);
                    cur=peek.right;
                }
            }
        }
        return res;
    }
}

2.2二叉树的层序遍历

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {

        List<List<Integer>> res = new ArrayList();
        // 用队列进行层序遍历
        Queue<TreeNode> queue = new LinkedList();
        queue.offer(root);
        if (root == null) {
            return res;
        }
        while (!queue.isEmpty()) {
            List<Integer> c = new ArrayList();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode poll = queue.poll();
                c.add(poll.val);
                if (poll.left != null) {
                    queue.offer(poll.left);
                    // c.add(poll.left.val);
                }
                if (poll.right != null) {
                    queue.offer(poll.right);
                    // c.add(poll.right.val);
                }
            }
            res.add(c);
        }
        return res;
    }
}