BM25 二叉树的后序遍历

https://www.nowcoder.com/practice/1291064f4d5d4bdeaefbf0dd47d78541?tpId=295&tqId=2291301&ru=/exam/oj&qru=/ta/format-top101/question-ranking&sourceUrl=%2Fexam%2Foj%3Fpage%3D1%26tab%3D%25E7%25AE%2597%25E6%25B3%2595%25E7%25AF%2587%26topicId%3D295

后序遍历的关键在于，记录最后一次访问的节点，避免重复访问

后序遍历访问顺序：左节点-> 右节点-> 当前节点

Go

package main

import . "nc_tools"

/*

* type TreeNode struct {

* Val int

* Left *TreeNode

* Right *TreeNode

* }

*/

/**

* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可

*

*

* @param root TreeNode类

* @return int整型一维数组

*/

func postorderTraversal( root *TreeNode ) []int {

// write code here

rc := root

stack := make([]*TreeNode,0,100)

result := make([]int,0,100)

var prev *TreeNode

for rc != nil || len(stack) > 0{

if rc == nil{

rc = stack[len(stack)-1]

if rc.Right != nil && rc.Right != prev{

rc = rc.Right

 

}else{

result = append(result,rc.Val)

prev = rc

stack = stack[:len(stack)-1]

rc = nil

}

}

for rc != nil{

stack = append(stack,rc)

rc = rc.Left

}

}

return result

}

C++

/**

* struct TreeNode {

* int val;

* struct TreeNode *left;

* struct TreeNode *right;

* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}

* };

*/

class Solution {

public:

/**

* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可

*

*

* @param root TreeNode类

* @return int整型vector

*/

vector<int> postorderTraversal(TreeNode* root) {

// write code here

TreeNode* rc = root;

stack<TreeNode*> st;

vector<int> result;

TreeNode* prev = nullptr;

while(rc != nullptr || st.size() > 0){

if(rc == nullptr){

rc = st.top();

if(rc->right != nullptr && rc->right != prev){

rc = rc->right;

}else{

result.push_back(rc->val);

prev = rc;

st.pop();

rc = nullptr;

}

}

while(rc != nullptr){

st.push(rc);

rc = rc->left;

}

}

return result;

}

};