You are given the root of a binary tree where each node has a value 0 or 1. Each root-to-leaf path represents a binary number starting with the most significant bit.
For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.
For all leaves in the tree, consider the numbers represented by the path from the root to that leaf. Return the sum of these numbers.
The test cases are generated so that the answer fits in a 32-bits integer.
The number of nodes in the tree is in the range [1, 1000].
Node.val is 0 or 1.
Solutions
Solution 1: Recursion
We design a recursive function dfs(root, t), which takes two parameters: the current node root and the binary number corresponding to the parent node t. The return value of the function is the sum of the binary numbers represented by the path from the current node to the leaf node. The answer is dfs(root, 0).
The logic of the recursive function is as follows:
If the current node root is null, then return 0. Otherwise, calculate the binary number t corresponding to the current node, i.e., t = t << 1 | root.val.
If the current node is a leaf node, then return t. Otherwise, return the sum of dfs(root.left, t) and dfs(root.right, t).
The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the binary tree. Each node is visited once; the recursion stack requires $O(n)$ space.
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# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclassSolution:defsumRootToLeaf(self,root:TreeNode)->int:defdfs(root,t):ifrootisNone:return0t=(t<<1)|root.valifroot.leftisNoneandroot.rightisNone:returntreturndfs(root.left,t)+dfs(root.right,t)returndfs(root,0)
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcsumRootToLeaf(root*TreeNode)int{vardfsfunc(root*TreeNode,tint)intdfs=func(root*TreeNode,tint)int{ifroot==nil{return0}t=(t<<1)|root.Valifroot.Left==nil&&root.Right==nil{returnt}returndfs(root.Left,t)+dfs(root.Right,t)}returndfs(root,0)}