二叉树的遍历(Java版)

递归最简单，迭代要用栈（其实也是模拟递归）。

前序遍历

递归

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

迭代

class Solution {
    List<Integer> res = new ArrayList<Integer>();
    Stack<TreeNode> stack = new Stack<TreeNode>();
    public List<Integer> preorderTraversal(TreeNode root) {
        if(root == null) return res;
        //前序根节点直接访问
        stack.push(root);
        while(!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if(cur.right != null) {
                stack.push(cur.right);
            }

            if(cur.left != null) {
                stack.push(cur.left);
            }
        }
        return res;
    }
}

中序遍历

递归

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

迭代

class Solution {
    List<Integer> res = new ArrayList<Integer>();
    Stack<TreeNode> stack = new Stack<TreeNode>();
    public List<Integer> inorderTraversal(TreeNode root) {
        if(root == null) return res;
        //中序 左-根-右
        while(root != null || !stack.isEmpty()) {
            //根节点不空就直接压栈
            if(root != null) {
                stack.push(root);
                root = root.left;
            }else{
                //根节点为空出栈直接访问左节点
                root = stack.pop();
                res.add(root.val);
                root = root.right;
            }
        }
        return res;
    }
}

后序遍历

递归

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

迭代

class Solution {
    LinkedList<Integer> res = new LinkedList<Integer>();
    Stack<TreeNode> stack = new Stack<TreeNode>();

    public List<Integer> postorderTraversal(TreeNode root) {
        if(root == null) return res;
        //根节点最后访问先入栈
        stack.push(root);
        while(!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.addFirst(cur.val);
            if(cur.left != null) {
                stack.push(cur.left);
            }
            if(cur.right != null) {
                stack.push(cur.right);
            }
            
        }

        return res;
    }
}