Binary Tree Postorder Traversal

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Given a binary tree, return the postorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [3,2,1].

#include<iostream>
#include<vector>
#include<stack>
using namespace std;

struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}

};
vector<int> postorderTraversal(TreeNode *root) {
	vector<int>ResultVectorPsotorder;
	stack<TreeNode*>StackNode;
	if (root == NULL)
		return ResultVectorPsotorder;
	StackNode.push(root);
	TreeNode*CurNode = NULL;
	TreeNode* PreOutNode = NULL;
	while (!StackNode.empty()){
		CurNode = StackNode.top();
		if ((CurNode->right==NULL&&CurNode->left==NULL)||
			(PreOutNode!=NULL&&(PreOutNode == CurNode->left || PreOutNode == CurNode->right))){
			ResultVectorPsotorder.push_back(CurNode->val);
			StackNode.pop();
			PreOutNode = CurNode;
		}
		else{
			if (CurNode->right)
				StackNode.push(CurNode->right);
			if (CurNode->left)
				StackNode.push(CurNode->left);
		}
	}
	return ResultVectorPsotorder;
}

Binary Tree Postorder Traversal

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原文地址：http://blog.csdn.net/li_chihang/article/details/43562801