leetcode:Binary Tree Postorder Traversal

最新推荐文章于 2022-07-02 16:50:24 发布

martin_liang

最新推荐文章于 2022-07-02 16:50:24 发布

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本文介绍了一种不使用递归实现二叉树后序遍历的方法，通过辅助栈来保存节点并最终返回节点值的后序遍历序列。

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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].

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

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/**
 * 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 {
    
private:
    void pushNodeToStack(TreeNode* root, stack<TreeNode*> &auxStack) {
        
        TreeNode* curNode = root;
        while (curNode->left != NULL || curNode->right != NULL)
        {
            if (curNode->right)
                auxStack.push(curNode->right);
            if (curNode->left)
                auxStack.push(curNode->left);
                
            curNode = auxStack.top();
        }
        return;
    }
    
public:
    vector<int> postorderTraversal(TreeNode* root) {
        
        vector<int> retVtr;
        if (root == NULL)
            return retVtr;
            
        stack<TreeNode *> auxStack;
        auxStack.push(root);
        pushNodeToStack(root, auxStack);
        
        while (auxStack.size() > 0)
        {
            TreeNode *preNode = NULL;
            while (auxStack.size() > 0 && (preNode == auxStack.top()->left || preNode == auxStack.top()->right))
            {
                TreeNode *curNode = auxStack.top();
                preNode = curNode;
                auxStack.pop();
                retVtr.push_back(curNode->val);
            }
            if (auxStack.size() > 0)
                pushNodeToStack(auxStack.top(), auxStack);
        }
        
        return retVtr;
    }
};