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Iterative binary tree traversals
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vector<int> inorderTraversal(TreeNode* root) { | |
unordered_map<TreeNode*, int> count; | |
stack<TreeNode*> stk; | |
vector<int> traversal; | |
stk.push(root); | |
while(!stk.empty()) { | |
TreeNode* cur = stk.top(); | |
if(!cur) { | |
stk.pop(); | |
continue; | |
} | |
switch(count[cur]) { | |
case 0: | |
stk.push(cur->left); | |
break; | |
case 1: | |
traversal.push_back(cur->val); | |
break; | |
case 2: | |
stk.push(cur->right); | |
break; | |
default: | |
stk.pop(); | |
break; | |
} | |
++count[cur]; | |
} | |
return traversal; | |
} | |
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vector<int> inorderTraversal(TreeNode* root) { | |
stack<pair<TreeNode*, int>> stk; | |
vector<int> traversal; | |
stk.push({root, 0}); | |
while(!stk.empty()) { | |
auto [cur, state] = stk.top(); | |
stk.pop(); | |
if(!cur || state > 2) { | |
continue; | |
} | |
stk.push({cur, state + 1}); | |
switch(state) { | |
case 0: | |
stk.push({cur->left, 0}); | |
break; | |
case 1: | |
traversal.push_back(cur->val); | |
break; | |
case 2: | |
stk.push({cur->right, 0}); | |
break; | |
} | |
} | |
return traversal; | |
} |
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