https://leetcode.com/problems/binary-tree-postorder-traversal/ Given a binary tree, return thepostordertraversal of its nodes' values. Example: [1,null,2,3] [3,2,1] Follow up: Recursive solution is trivial, could you do it iteratively? 回到顶部 Intuition Method 1. Using one stack to s...
returngetpost(pl+1,pl+len,il,i-1); } else{//没有左子树,则插入右边 returngetpost(pl+len+1,pr,i+1,ir); } } intmain(){ intn,i,k; cin>>n; for(i=0;i<n;i++) cin>>pre[i]; for(i=0;i<n;i++) cin>>in[i]; cout<<getpost(0,n-1,0,n-1)<<endl; return0; } 1...
Given a binary tree, return thepostordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[3,2,1]. Note:Recursive solution is trivial, could you do it iteratively? ++++++++++++++++++++++++++++++++++++++++++ 【二叉树遍历模版】后...
Postorder traversal is 4 2 7 8 5 6 3 1 Die Zeitkomplexität der obigen Lösung istO(n), wonist die Gesamtzahl der Knoten im Binärbaum. Das Programm erfordertO(n)zusätzlicher Platz für die Hash-Tabelle und den Call-Stack.Hierist eine Implementierung in C, die in ausgeführt...
// Pushe das linke und rechte Kind des gepoppten Knotens in den Stack if (curr->left) { s.push(curr->left); } if (curr->right) { s.push(curr->right); } } // Postorder-Traversal drucken while (!out.empty()) { cout << out.top() << " "; out.pop(); } } int main(...
This space is primarily used for the recursive call stack during the construction of the tree. Additionally, the inorder_index dictionary requires O(n) space, as it stores the index of each value in the inorder traversal list. Overall, the space usage is proportional to the number of nodes...
root->r=bt(arr,mid+1,last); returnroot; } /* Insert a node in the tree */ voidinsert() { intarr1[]={10,30,20,90,80,60,40},i; printf("Given post order traversal array\n"); for(i=0;i<=6;i++) { printf("%d->",arr1[i]); ...
function postorderTraversal(root) { if (!root) return []; const stack = [root]; const result = []; while (stack.length) { const node = stack[stack.length - 1]; if (node.left) { stack.push(node.left); node.left = null; } else if (node.right) { stack.pu...
For each test case, print in one line the first number of the postorder traversal sequence of the corresponding binary tree. Sample Input: 7 1 2 3 4 5 6 7 2 3 1 5 4 7 6 Sample Output: 3 坑点: 1. 唯一要注意的是递归太深,会导致堆栈溢出,所以要控制递归返回条件 ...
1. A binary tree cannot be reverted to the only binary tree by using the sequence of preorder traversal,inorder traversal,postorder traversal or Node-Degree. 用二叉树的前序遍历、中序遍历、后序遍历的序列或结点度表示法都无法还原为唯一的一棵二叉树,中序遍历和结点度表示法二者结合组成一个序列,此...