二叉树的后序遍历
用标记右子树vector的方法
vector<int> postorderTraversal(TreeNode *root) {
vector<int> ans;
vector<TreeNode *> stack;
vector<bool> isRight;
stack.push_back(root);
isRight.push_back(false);
TreeNode * pNode = NULL;
while(!stack.empty())
{
while((pNode = stack.back()) != NULL)
{
pNode = pNode->left;
stack.push_back(pNode);
isRight.push_back(false);
} while(isRight.back() == true)
{
stack.pop_back();
isRight.pop_back();
ans.push_back(stack.back()->val);
}
stack.pop_back();
isRight.pop_back(); if(!stack.empty())
{
pNode = stack.back();
stack.push_back(pNode->right);
isRight.push_back(true);
}
} return ans;
}
仅用一个栈的方法https://oj.leetcode.com/discuss/14118/post-order-traversal-using-two-satcks
vector<int> postorderTraversal(TreeNode *root) {
stack<TreeNode *> st;
vector<int> vRet;
TreeNode *p, *pre = root; if (!root) return vRet;
p = root->left;
st.push(root);
while (!st.empty() )
{
while (p) { st.push(p); p = p->left; }
p = st.top();
while (!p->right || pre==p->right)
{
vRet.push_back(p->val);
pre = p;
st.pop();
if (st.empty() ) break;
p = st.top();
}
p = p->right;
}
return vRet;
}