Created
May 5, 2023 12:57
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Inorder traversal of binary tree, from recursive to iterative
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const tree = { | |
value: 'A', | |
left: { | |
value: 'B', | |
left: { | |
value: 'D', | |
left: { | |
value: 'H', | |
}, | |
right: { | |
value: 'I' | |
}, | |
}, | |
right: { | |
value: 'E', | |
left: { | |
value: 'J' | |
}, | |
}, | |
}, | |
right: { | |
value: 'C', | |
left: { | |
value: 'F', | |
}, | |
right: { | |
value: 'G', | |
}, | |
}, | |
}; | |
// recursive | |
function inOrderTraverse1(root) { | |
if (!root) { | |
return; | |
} | |
inOrderTraverse1(root.left); | |
console.log(root.value); | |
inOrderTraverse1(root.right); | |
} | |
// call = push stack variables and return address to stack, setup stack variables according to function parameter, then goto beginning | |
// return = pop stack variables from stack, reset stack variables and execution address from stack content, then and go to beginning | |
// given that we only have `root` as stack variable, the stack is just an array of roots | |
function inOrderTraverse2(root) { | |
const stack = []; | |
let pc = 0; | |
while (true) { | |
switch (pc) { | |
case 0: { | |
if (!root) { | |
// return, i.e. pop stack | |
if (stack.length === 0) { | |
return; | |
} | |
({ root, pc } = stack.pop()); | |
continue; | |
} | |
// inOrderTraverse1(root.left); | |
stack.push({ root, pc: 1 }); | |
root = root.left; | |
pc = 0; | |
continue; | |
} | |
case 1: { | |
console.log(root.value); | |
// inOrderTraverse1(root.right); | |
stack.push({ root, pc: 2 }); | |
root = root.right; | |
pc = 0; | |
continue; | |
} | |
case 2: { | |
// return, i.e. pop stack | |
if (stack.length === 0) { | |
return; | |
} | |
({ root, pc } = stack.pop()); | |
continue; | |
} | |
} | |
} | |
} |
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