Minimum Depth of Binary Tree (E)
题目
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
Note: A leaf is a node with no children.
Example:
Given binary tree [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
return its minimum depth = 2.
题意
找到给定树从根到叶结点(两子树都为空的结点)的最短距离。
思路
递归或迭代。
代码实现
Java
递归
class Solution {
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
int lDepth = minDepth(root.left);
int rDepth = minDepth(root.right);
if (lDepth == 0) {
return rDepth + 1;
} else if (rDepth == 0) {
return lDepth + 1;
} else {
return Math.min(lDepth, rDepth) + 1;
}
}
}
迭代
class Solution {
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
Queue<TreeNode> q = new ArrayDeque<>();
int depth = 0;
q.offer(root);
while (!q.isEmpty()) {
depth++;
int size = q.size();
for (int i = 0; i < size; i++) {
TreeNode cur = q.poll();
// 第一次找到叶结点，说明已经为最短距离
if (cur.left == null && cur.right == null) {
return depth;
}
if (cur.left != null) {
q.offer(cur.left);
}
if (cur.right != null) {
q.offer(cur.right);
}
}
}
return depth;
}
}
JavaScript
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var minDepth = function (root) {
if (!root) {
return 0
}
return (
(!root.left
? minDepth(root.right)
: !root.right
? minDepth(root.left)
: Math.min(minDepth(root.left), minDepth(root.right))) + 1
)
}
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