Leetcode 145 binary Tree Postorder Traversal (subsequent traversal of binary trees) + (binary tree, iteration)

translation

给定一个二叉树，返回其后续遍历的节点的值。例如：给定二叉树为 {1#， 2， 3}   1         2    /   3返回 [321]备注：用递归是微不足道的，你可以用迭代来完成它吗？

Original

returntheofits nodes‘ values.For example:Given binary tree {1,#,2,3},   1         2    /   3return [3,2,1isit iteratively?

Analysis

Directly on the code ...

vector<int> postorderTraversal(TreeNode* root) {    if (root != NULL) {        postorderTraversal(root->left);        postorderTraversal(root->right);             v.push_back(root->val);    }    return v;}

/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode (int x): Val (x), left (null), right (null) {}*};*/classSolution { Public: vector<int>VvoidPostordertraversaliter (TreeNode *root, Stack<TreeNode*>&AMP;STAC) {if(Root = NULL)return;BOOLHasleft = root->left! = NULL;BOOLHasright = root->right! = NULL; Stac.push (root);if(hasright) Stac.push (root->right);if(hasleft) Stac.push (root->left);if(!hasleft &&!hasright) v.push_back (root->val);if(Hasleft)            {root = Stac.top ();            Stac.pop ();        Postordertraversaliter (Root, Stac); }if(Hasright)            {root = Stac.top ();            Stac.pop ();        Postordertraversaliter (Root, Stac); }if(Hasleft | | hasright) v.push_back (Stac.top ()->val);    Stac.pop (); } vector<int>Postordertraversal (treenode* root) { Stack<TreeNode*>Stac Postordertraversaliter (Root, Stac);returnV }     };

There are two other similar topics:

Leetcode 94 binary Tree Inorder traversal (middle order traversal of binary trees) + (binary tree, iteration)
Leetcode 144 binary Tree preorder traversal (binary trees pre-sequence traversal) + (binary tree, iteration)

Leetcode 145 binary Tree Postorder Traversal (subsequent traversal of binary trees) + (binary tree, iteration)