binary-tree-postorder-traversal——二叉树后续遍历

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].

Note: Recursive solution is trivial, could you do it iteratively?

递归方法

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

非递归

处理并出栈时判断，该节点是否有子节点或其子节点是否被访问过。并记录上次处理的节点。

/**
 * 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) {
        vector<int> v;
        if(root==NULL)
            return v;
        stack<TreeNode *> tree;
        tree.push(root);
        TreeNode *pre=NULL;
        while(!tree.empty()){
            TreeNode *p = tree.top();
            if((p->left==NULL&&p->right==NULL)||(pre!=NULL&&(p->left==pre||p->right==pre))){
                v.push_back(p->val);
                pre=p;
                tree.pop();
            }else{
                if(p->right!=NULL)
                    tree.push(p->right);
                if(p->left!=NULL)
                    tree.push(p->left);
            }
        }
        return v;
    }
};