Binary Tree Postorder Traversal

最新推荐文章于 2022-07-19 11:52:33 发布

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最新推荐文章于 2022-07-19 11:52:33 发布

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本文链接：https://blog.youkuaiyun.com/zzyafyj/article/details/41520407

本文介绍了如何使用递归和非递归方式实现二叉树的后序遍历，包括定义了二叉树节点结构，并提供了具体的实现代码。

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

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

return [3,2,1].

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 
class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        if(root){
			postorderTraversal(root->left);
			postorderTraversal(root->right);
			values.push_back(root->val);
			return values;
		}else{
		    return values;
		}
    }
private:
	vector<int> values;
};

。。。我就说怎么会有这么简单的题，没看到用非递归

非递归:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 
class Solution {
public:
	vector<int> postorderTraversal(TreeNode *root) {
		 stack<TreeNode *> paths;
		 map<TreeNode *,bool> visited;
		 while(root || !paths.empty()){
			 while(root){
			      paths.push(root);
				  root=root->left;
			 }
			 if(!paths.empty()){
				 TreeNode *p= paths.top();
			     while(!paths.empty() && visited[p]){//p的右子树已经访问过了
					 values.push_back(p->val);
					 paths.pop();
					 if(!paths.empty())
					      p=paths.top();
				 }
			     if(!visited[p]){
					 visited[p]=true;
					 root=p->right;
				 };
			 }
		 }
		 return values;
	}
private:
	vector<int> values;
};