二叉树的深度优先遍历

结合代码随想录和力扣大佬部分解析programmercarl.com/%E4%BA%8C%E… programmercarl.com/%E4%BA%8C%E… leetcode.cn/problems/bi…

1. 144. 二叉树的前序遍历

• 递归

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

• 迭代

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null){
            return res;
        }
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);//入栈顺序：中-右-左 出栈顺序：中-左-右
        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;
    }
}

2. 94. 二叉树的中序遍历

• 递归

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

• 迭代

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

3. 145. 二叉树的后序遍历

• 递归

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

• 迭代

    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null){
            return res;
        }
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);//入栈顺序：中-左-右 出栈顺序：中-右-左， 最后翻转结果
        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;
    }
}