【leetcode刷題】145. Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [3,2,1]

Follow up: Recursive solution is trivial, could you do it iteratively?

 

class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
 
import java.util.List;
import java.util.ArrayList;
import java.util.Stack;
class Solution {
 public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null) return res;
        postDfs(root, res);
        return res;
    }
    public void postDfs(TreeNode root, List<Integer> res){
        if(root == null) return;
        postDfs(root.left, res);
        postDfs(root.right, res);
        res.add(root.val);
    }
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null) return res;
        TreeNode p = root, r = null;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        while(p != null || !stack.isEmpty()){
            if(p != null){   //樹左邊的元素入棧
                stack.push(p);
                p = p.left;
            }else{
                p = stack.peek(); 
                if(p.right != null && p.right != r){ //右子樹存在，未被訪問
                    p = p.right;
                }else{
                    stack.pop();
                    res.add(p.val);
                    r = p;//記錄最近訪問過的節點
                    p = null;//節點訪問完後，重置p指針
                }
            }
        }
        return res;
    }
}

import java.util.List;
import java.util.ArrayList;
import java.util.Stack;
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
 }

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