145. Binary Tree Postorder Traversal

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

For example:
Given binary tree [1,null,2,3],

   1\2/3

思路： 借助于一个栈，依次将根节点的右子节点和左子节点压入栈中。如果一个节点为叶子节点，或者前一个出栈的元素为当前栈顶节点的子节点，则出栈。

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

转载于:https://www.cnblogs.com/zhaoyaxing/p/8502683.html

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