翻译

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 [3, 2, 1]备注：用递归是微不足道的，你可以用迭代来完成它吗？

原文

Given a binary tree, return the postorder traversal of its nodes' values.For example:Given binary tree {
    1,#,2,3},   1    \     2    /   3return [3,2,1].Note: Recursive solution is trivial, could you do it iteratively?

分析

直接上代码……

vector
     
       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) {}* };*/class Solution {public:         vector
     
       v;    void  postorderTraversalIter(TreeNode *root, stack
      
        &stac) {        if (root == NULL) return;        bool hasLeft = root->left != NULL;        bool hasRight = 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
       
         postorderTraversal(TreeNode* root) {        stack
        
          stac;        postorderTraversalIter(root, stac);        return v;    }     };
        
       
      
     

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