二叉树的前中后 递归与非递归遍历

一、前序

• 递归

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

• 非递归

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
       List<Integer> res = new ArrayList<>();
       Stack<TreeNode> stk = new Stack<>();

       if(root == null)return res;
       stk.push(root);
       while(!stk.isEmpty()){
            TreeNode t = stk.pop();
            if(t != null){
                if(t.right != null)stk.push(t.right);
                if(t.left != null)stk.push(t.left);
            }
            res.add(t.val);
        }
        return res;
    }
}

二、后序

• 递归

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

• 非递归

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Stack<TreeNode> stk = new Stack<>();
        Stack<Integer> help = new Stack<>();
        if(root == null)return res;
        stk.push(root);
        while(!stk.isEmpty()){
            TreeNode t = stk.pop();
            if(t != null){
                if(t.left != null)stk.push(t.left);
                if(t.right != null)stk.push(t.right);
            }
            // System.out.println(t.val);
            help.push(t.val);
        }
        while(!help.isEmpty())res.add(help.pop());
        return res;
    }   
}

三、中序

• 递归

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

• 非递归

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Stack<TreeNode> stk = new Stack<>();
        TreeNode cur = root;
        if(cur == null)return res;

        while(cur != null || !stk.isEmpty()){
            if(cur != null){
                stk.push(cur);
                cur = cur.left;
            }else{
                TreeNode t = stk.pop();
                res.add(t.val);
                cur = t.right;
            }
        }
        return res;
    }
}