Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree
{1,#,2,3}
,1 \ 2 / 3return
[1,3,2]
.
A:
-----用stack迭代 , 每个节点pop出来的时候,才去visit, 入栈时走左边,出栈后,走右边
又是双100%, LOL
/** * 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 { public: vector<int> inorderTraversal(TreeNode* root) { vector<int> res; vector<TreeNode*> stack; while(root || !stack.empty()){ if(root){ stack.push_back(root); root=root->left; }else{ root = stack.back(); stack.pop_back(); res.push_back(root->val); root = root->right; } } return res; } };
Learned:
1: in-order iterative traversal的关键点是两个:
1) 把左节点先都加进去, 然后, 一个个地弄出来,visit + 考虑右子树
2) 在循环的时候,并不是每次都pop,
3) 通过curNode 是否为null来标记 当前我们是否走到了right节点的最下边,不是的话,就要重新考虑其左右子节点。 (或者说,标记是否刚刚走的是right 节点)
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