1. 二叉树的后序遍历 - Binary Tree Postorder Traversal

145. 二叉树的后序遍历 - Binary Tree Postorder Traversal

【后序遍历-递归】【后序遍历-迭代】【后序遍历-先序的变异】

Problem Link

给定一个二叉树,返回它的 后序 遍历。题目描述

Example:

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

输出: [3,2,1]

Solutions

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

Solution 【后序遍历-递归】 ( 0ms)

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class Solution {
    
    List<Integer> result =new ArrayList<Integer>();
    
    public List<Integer> postorderTraversal(TreeNode root) {
        loop(root);
        return result;
    }
    /* 单独写递归循环,减少return的次数 */
    public void loop(TreeNode node){
        if(node==null)
            return;
        loop(node.left);
        loop(node.right);
        result.add(node.val);
    }
}

Solution 【后序遍历-迭代】 ( 2ms)

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class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        if(root == null)
            return result;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode pre = null;
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode curr = stack.peek();            
            if((curr.left == null && curr.right == null) ||
               (pre != null && (pre == curr.left || pre == curr.right))){
                result.add(curr.val);
                pre = curr;
                stack.pop();
            }else{/* 先将右结点压栈, 再将左结点入栈 */
                if(curr.right != null) 
                    stack.push(curr.right); 
                if(curr.left != null) 
                    stack.push(curr.left);   
            }            
        }
        return result;        
    }
}

Solution 【后序遍历-先序的变异】 ( 2ms)

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class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        if(root == null)
            return result;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            if(node.left != null) stack.push(node.left);//和传统先序遍历不一样,先将左结点入栈
            if(node.right != null) stack.push(node.right);//后将右结点入栈
            result.add(0,node.val);                        //逆序添加结点值
        }     
        return result;
    }
}

Notes

updatedupdated2021-03-052021-03-05