0404.左叶子之和
题目大意
求所有左叶子节点的值的和
思路与代码
深度优先遍历
/**
* 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 sumOfLeftLeaves = function (root) {
let sum = 0;
const dfs = (node) => {
if (!node) return;
const left = node.left;
if (left && !left.left && !left.right) {
sum += left.val;
}
dfs(node.left);
dfs(node.right);
};
dfs(root);
return sum;
};空间复杂度:O(h)
时间复杂度:O(n)
广度优先遍历
/**
* 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 sumOfLeftLeaves = function (root) {
if (!root) return 0;
let sum = 0;
const queue = [root];
while (queue.length) {
const n = queue.length;
for (let i = 0; i < n; i++) {
const node = queue.shift();
if (node.left && !node.left.left && !node.left.right) {
sum += node.left.val;
}
if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);
}
}
return sum;
};空间复杂度:O(n)
时间复杂度:O(n)
