How to parse the minimum depth of python binary tree
How to analyze the minimum depth of python binary tree, I believe that many inexperienced people do not know what to do. Therefore, this paper summarizes the causes and solutions of the problem. Through this article, I hope you can solve this problem.
Topic description
Given a binary tree, find out its minimum depth.
The minimum depth is the number of nodes on the shortest path from the root node to the nearest leaf node.
Note: a leaf node is a node that does not have child nodes.
Example:
Given binary tree [3pc9pr 20pr nullpr nullpr 15pc7]
three
/\
9 20
/\
15 7
Returns its minimum depth 2.
The idea of solving the problem
Tag: DFS
Termination conditions, return values, and recursive procedures:
If the root of the current node is empty, the height of the tree here is 0 and 0 is also the minimum.
When both the left and right subtrees of the current node root are empty, it means that the height of the tree here is 1 and 1 is also the minimum
If it is otherwise, it means that the current node has a value, and the minimum depth of its left and right subtrees needs to be calculated respectively. Returning the minimum depth + 1 means that there is 1 depth in the current node.
Time complexity: O (n), n is the number of nodes in the tree
Code
Java version
/ * *
* Definition for a binary tree node.
* public class TreeNode {
* int val
* TreeNode left
* TreeNode right
* TreeNode (int x) {val = x;}
*}
, /
Class Solution {
Public int minDepth (TreeNode root) {
If (root = = null) {
Return 0
}
If (root.left = = null & & root.right = = null) {
Return 1
}
Int ans = Integer.MAX_VALUE
If (root.left! = null) {
Ans = Math.min (minDepth (root.left), ans)
}
If (root.right! = null) {
Ans = Math.min (minDepth (root.right), ans)
}
Return ans + 1
}
}
JavaScript version
/ * *
* Definition for a binary tree node.
* function TreeNode (val) {
* this.val = val
* this.left = this.right = null
*}
, /
/ * *
* @ param {TreeNode} root
* @ return {number}
, /
Var minDepth = function (root) {
If (root = = null) {
Return 0
}
If (root.left = = null & & root.right = = null) {
Return 1
}
Let ans = Number.MAX_SAFE_INTEGER
If (root.left! = null) {
Ans = Math.min (minDepth (root.left), ans)
}
If (root.right! = null) {
Ans = Math.min (minDepth (root.right), ans)
}
Return ans + 1
}
Drawing and interpretation
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